Chapter 3
Population Genetics:
Difference Equation Models
The diversity of life is a fundamental empirical fact of nature. Casual observation alone confirms the great variety of species. But diversity also prevails
within single species. As in human populations, the individuals of a species
vary considerably in their outward traits—size and shape, external markings, fecundity, disease resistance, etc. This is called phenotypic variation,
and, since phenotypes are shaped by genes, it reflects an underlying, genetic
diversity. The achievements of genetics and molecular biology, as described
in Chapter 1, have allowed us to measure and confirm the genotypic variability of populations down to the molecular level.
The science of genotypic variation in interbreeding populations is called
population genetics. Its goal is understanding how genetic variation changes
under the influence of selection, mutation, and the randomness inherent in
mating, as one generation succeeds another. It is a quantitative subject in
which mathematical modeling and analysis play key roles. Population geneticists measure the genetic composition of a population by the frequencies
of the various genotypes that are present. They combine what is known
about the mechanisms of heredity—how DNA carries genetic information,
how chromosomes function and give rise to Mendel’s laws, how mutations
arise—with hypotheses about mating and selective advantage, to propose
mathematical equations for the evolution of genotype frequencies. They
then analyze the solutions of these equations to derive predictions that can
be used to test hypotheses on mating and selection, to understand how
different factors affect genotype variation, and to make inferences about
genealogy and evolution.
This chapter is an introduction to elementary mathematical models of
population genetics for large populations. These models are derived by
imagining how genotype frequencies evolve in the limit as the population
1
2
CHAPTER 3. POPULATION GENETICS I
size tends to infinity. The use of this limit is called the “infinite population
assumption,” and it is imposed throughout the chapter. One of its main
consequences is that genotype frequencies become deterministic functions
of time, and hence infinite population models take the form of deterministic dynamical systems. However, the rules governing the dynamics of the
frequencies —how they change in time—are derived by probabilistic reasoning, and, to understand the chapter, the reader should review the material
in Section 2.1 on random sampling, independence, the law of large numbers,
and conditional probability.
Population genetics is perhaps the most prominent field of biology for
applications of probability. But this is not the only reason it is treated first
in this book. It also makes an excellent introduction to probabilistic modeling in general. In reading this chapter, the student should study not just
the models themselves, but also the process by which they are derived. To
help in this regard, Section 3.2 presents a careful discussion of the biological and probabilistic principles of population genetics. The presentation
of the models themselves in Section 3.3 follows a common strategy used in
applications of mathematics to science: begin with oversimplified models
and gradually derive more complex and realistic ones by relaxing different
assumptions one by one. Mathematical modeling is an art that can only be
learned by practice. Throughout, guided exercises in extending and developing the material are worked into the text, and the student should do them
as an integral part of reading the chapter.
The models derived in Sections 3.3 and 3.4 almost all take the form of
finite difference equations. Section 3.1 treats the essential theory of first and
second order, linear difference equations needed for Section 3.3. A graphical
method called cobwebbing is presented in Section 3.4 for analyzing nonlinear difference equations, in preparation for studying models with selection.
Difference equations have an importance extending far beyond population
genetics, since they arise as models in many applied areas.
3.1
Finite Difference Equations; Introduction
This section supplies an introduction to finite difference equations sufficient
for understanding Chapter 3. The reader can begin directly with the population genetics in section 3.2 and 3.3, and refer back to this section as the
need arises, but it is helpful to read at least the beginning of this section
first, to learn what difference equations are about.
Difference equations are recursive equations that determine a sequence of
numbers. It is common in mathematical texts to denote a generic sequence
by the notation (x1 , x2 , x3 , . . . , xn , . . .). In this chapter, we shall instead
denote a generic sequence by (x(0), x(1), x(2) . . .), and a generic term in the
3.1. FINITE DIFFERENCE EQUATIONS
3
sequence by x(t). We use t as the index because it will always represent
some sort of time parameter, and writing x(t) rather than xt will allow us
to use subscripts for other purposes. In general, our convention is that the
first element of a sequence denotes a value at time t = 0, and that is why
the first term in a sequence is taken to be x(0). It will also be convenenient
to abbreviate the sequence (x(0), x(1), . . .) by (x(t))t≥0 .
3.1.1
First-order difference equations
The general, autonomous, first-order difference equation for a sequence
(x(t))t≥0 takes the form
x(t+1) = φ(x(t)),
t ≥ 0,
(3.1)
where φ is a given function. We call (3.1) autonomous because φ(x) depends explicitly only on x. (The non-autonomous generalization of (3.1)
is x(t + 1) = φ(t, x(t)), where φ(t, x) depends on both t and x. We shall
encounter only autonomous equations, and so we restrict the discussion to
this case.) We call (3.1) first-order equation because, for each t, the value of
x(t+1) is determined completely by the immediately preceding value x(t).
Second-order difference equations will be defined and discussed in the next
subsection.
A specification,
x(0) = x0 ,
(3.2)
of a value for the first term is called an initial condition. Given a function φ
and an initial value x0 , there will be a unique sequence (x(t))t≥0 , called the
solution of (3.1)-(3.2), for which x(0) = x0 and x(t+1) = φ(x(t)) for all t ≥ 0.
This is easy to see by applying the difference equation for successively greater
values of t, starting from t = 0. Thus, setting t = 0 in (3.1) implies x(1) =
φ(x(0)) = φ(x0 ); then setting t = 1 implies x(2) = φ(x(1)) = φ(φ(x0 )); and,
continuing in this manner, x(t) = φ (φ(· · · φ(x0 ) · · ·)), where φ is convolved
with itself t times. By computing x(0), x(1), x(2), . . . recursively in this way,
one can calculate x(t) numerically for any desired t, so long as each new
value of x(t) is in the domain of the function φ. If at some point, x(t) is
not in the domain of φ, the procedure must stop and the solution to the
difference equation will be only a finite sequence.
Example 3.1.1. For a very simple example, consider the equation
1
x(t+1) = x(t),
2
x(0) = x0 .
We have, x(1) = x(0)/2 = x0 /2, x(2) = x(1)/2 = x0 /22 , and, dividing by
two at each step, x(t) = x0 /2t for any non-negative integer t.
4
CHAPTER 3. POPULATION GENETICS I
Example 3.1.2. Consider
x(t+1) =
x(t)
1
+
,
2
x(t)
x(0) = 1.
(3.3)
Then x(1) = 1/2 + 1 = 3/2 and x(2) = (3/2)/2 + 2/3 = 17/12 = 1.41666 . . .,
and so forth. We do not know an explicit, closed form expression for the
general term x(t). Observe that x = 0 is not in the domain of x/2 + 1/x.
Hence, if x(0) = 0 is the initial condition, the solution stops with this one
value and cannot continue for t ≥ 1. (Some authors would say that a solution
to (3.3) does not exist for x(0) = 0.)
For the purposes of this chapter, we need techniques from difference
equation theory to address two issues: finding, if possible, closed form solutions, and analyzing the behavior of solutions (x(t))t≥0 as t → ∞. The
latter problem is important because knowing the exact solution is often less
interesting than understanding whether or not x(t) settles into some type
of regular behavior as t increases. In particular, one often wants to know if
limt→∞ x(t) exists, and, if it does, what its value is. In the rest of section
3.1, we show how to resolve these questions for linear difference equations.
A first order difference equation is called linear if φ is linear; its general
form is
x(t+1) = αx(t) + β.
(3.4)
Such equations are particularly nice because they have simple, explicit solutions whose limiting behavior can be described completely.
Proposition 1 If α 6= 1, the general solution to (3.4) is
x(t) = Aαt +
β
, t ≥ 0, where A is an arbitrary constant,
1−α
(3.5)
and the solution to (3.4) satisfying the initial condition x(0) = x0 is
x(t) = x0 −
β
β
αt +
,
1−α
1−α
t ≥ 0.
(3.6)
If α = 1, the general solution to (3.4) is x(t) = A + βt, t ≥ 0, and the
solution with initial condition x(0) = x0 is x(t) = x0 + βt, t ≥ 0.
β
Assume α 6= 1. In Proposition 1, the claim that x(t) = Aαt + 1−α
, t ≥ 0,
for A arbitrary, is the general solution to (3.4) means that: (a) for any
β
number A, x(t) = Aαt + 1−α
, t ≥ 0, is a solution of (3.4); and (b),
conversely, any solution to (3.4) must have this form for some A. Let us
3.1. FINITE DIFFERENCE EQUATIONS
5
show why this must be true. To prove (a) we must demonstrate that (3.4)
β
is true when x(t) is replaced by Aαt + 1−α
; that is, we must show
Aα
t+1
β
β
+
= α Aαt +
+ β.
1−α
1−α
But this can be verified (do so!) by some simple algebra.
Now suppose the initial condition x(0) = x0 is specified. We can find a
solution of the form in (3.5) if we can find A such that
x0 = x(0) = Aα0 +
β
β
=A+
.
1−α
1−α
But this equation has the unique solution A = x0 − β/(1 − α) for any x0 ,
and using this A we obtain the solution (3.6).
To prove (b), let (z(t))t≥0 be any solution to (3.4). Then, from what we
have just shown,
β
β
y(t) = z(0) −
αt +
,
1−α
1−α
t ≥ 0,
is a solution with the same initial condition. Since solutions are uniquely
specified once initial conditions are specified, y(t) = z(t) for all t, and so
(z(t))t≥0 has the form z(t) = Aαt + β/(1 − α) with A = z(0) − β/(1 − α).
The statements in Proposition 1 about the solution when α = 1 are all
demonstrated in a similar manner.
It is interesting to analyze the solutions (3.5) and (3.6) further. Again,
consider the case α 6= 1. First, notice that setting A = 0 in (3.5) yields a
constant solution to (3.5), namely
x(t) =
β
,
1−α
for all t = 0, 1, 2, . . ..
This is called a fixed point solution of the difference equation. In fact, it is
the only constant solution, and can be found directly. Indeed, let (z(t))t≥0
be the constant sequence: z(t) = x̄ for all t ≥ 0. This will solve (3.5) if and
only if
x̄ = z(t+1) = αz(t) + β = αx̄ + β.
(3.7)
This has the unique solution x̄ = β/(1 − α), which therefore defines the
unique, fixed-point solution. In the case α = 1, if β 6= 0, x̄ = x̄ + β admits
no solution and hence x(t+1) = x(t)+β has no fixed point.
Second, the formulas (3.5) and (3.6) reveal completely how solutions
to (3.4) behave in the long time limit. If α 6= 1 there are two different
6
CHAPTER 3. POPULATION GENETICS I
limiting behaviors, according to whether |α| < 1 or |α| > 1. If |α| < 1, then
limt→∞ αt = 0, and hence
lim x(t) = x0 −
t→∞
β
β
β
lim αt +
=
.
1 − α t→∞
1−α
1−α
(3.8)
In other words, as t increases to ∞, the solution x(t) tends to the fixed point
β/(1 − α), no matter what the initial value is. One describes this situation
by saying that the fixed point is globally asymptotically stable. On the other
hand, if |α| > 1, the sequence {|α|t ; t ≥ 0} goes off to ∞ as t → ∞, and
hence for any x0 not equal to the fixed point, |x(t)| → ∞. In this case,
β/(1−α) is an example of an unstable fixed point.
Turning to the case α = 1, one sees immediately from (3.7) that x(t+1) =
x(t) + β admits no fixed point unless β = 0, in which case the fixed point is
x̄ = 0. Because the general solution is A + βt, solutions will tend to +∞ or
−∞, according to whether β > 0 or β < 0; when β = 0 solutions are always
constant.
These limiting results, although derived for linear difference equations,
illustrate concepts that generalize to the nonlinear case. A point x̄ is called
a fixed point of the general difference equation x(t+1) = φ(t) if φ(x̄) = x̄.
The sequence which takes the constant value x̄ therefore solves the difference
equation and is called a fixed-point solution. In general, a difference equation
may have many different fixed points. Now suppose that φ is a continuous
function and (x(t))t≥0 is a solution to x(t + 1) = φ(x(t)) such that x̄ =
limt→∞ x(t) exists and is finite. Then, using the continuity of φ,
x̄ = lim x(t + 1) = lim φ(x(t)) = φ( lim x(t)) = φ(x̄).
t→∞
t→∞
t→∞
Thus the limit x̄ must be a fixed point of the equation. This simple fact
at least identifies all potential candidates for limits of solutions. If x̄ is a
fixed point of φ, define the basin of attraction of x̄ to be the set of points
x0 such that the solution of x(t+1) = φ(t) with initial condition x(0) = x0
converges to x̄. For example, in the case of linear the linear difference
equation (3.5) with α 6= 1, we saw that if |α| < 1, all solutions tend to the
fixed point β/(1 − α), and therefore its basin of attraction is the set of all
points in (−∞, ∞). On the other hand, if |α| > 1, the basin of attraction
consists solely of the fixed point. A fixed point is called stable if its basin
of attraction contains an open interval about x0 . For nonlinear equations,
basins of attraction are in general difficult to calculate exactly. However, a
simple criterion for stability is presented in section 3.4.
3.1. FINITE DIFFERENCE EQUATIONS
3.1.2
7
Second-order, linear difference equations.
A second-order (autonomous) difference equation takes the general form:
x(t+1) = ψ (x(t), x(t−1)) ,
t ≥ 1.
(3.9)
Now, each term in the sequence is determined by the two previous values,
and in order to start a solution off, it is necessary to specify initial values of
x(0) and x(1):
x(0) = x0 ,
x(1) = x1 .
(3.10)
Equations (3.9)–(3.10) have a unique solution which can be computed numerically by recursion: thus, x(2) = φ(x1 , x0 ), x(3) = φ(x(2), x(1)) =
φ(φ(x1 , x0 ), x1 ), etc., and the recursion continues so long as (x(t), x(t−1))
remains in the domain of φ.
We shall encounter population genetics models which are second-order,
linear difference equations, that is, are of the form:
x(t+1) = αx(t) + γx(t−1) + β.
(3.11)
This equation can be solved by a method reminiscent of the theory of secondorder, linear differential equations. Consider first the case in which β = 0:
x(t+1) = αx(t) + γx(t−1),
(3.12)
(where γ 6= 0, so that the equation is truly second-order). This is called the
homogeneous case.
The virtue of the homogeneous equation is that solutions to it obey a
superpostion principle: if (z(t))t≥0 and (y(t))t≥0 both solve (3.12), then so
does the linear combination (x(t) = Az(t) + By(t)), for any constants A and
B. Indeed, if z(t+1) = αz(t) + γz(t−1) and y(t+1) = αy(t) + γy(t−1), then
x(t+1) = Az(t+1) + By(t+1) = A [αz(t) + γz(t−1)] + B [y(t) + γy(t−1)]
= α [Az(t) + By(t)] + γ [Az(t−1) + By(t−1)]
= αx(t) + γx(t−1),
and thus (x(t))t≥0 solves (3.12) as well. This superposition principle is the
key to the solution method. It turns out that if (z(t))t≥0 and (y(t))t≥0 are
independent solutions of (3.12), in the sense that (y(t))t≥0 is not a constant
multiple of (z(t))t≥0 , then Az(t) + By(t), t ≥ 0, is the general solution of
(3.12); that is, all solutions of (3.12) have this form. Suppose we now want
to solve (3.12), subject to the initial conditions x(1) = x1 and x(0) = x0 .
We know it must be be x(t) = Az(t) + By(t) for some A and B, and to find
A and B, we need only to solve the equations for A and B required by the
initial conditions:
Az(0) + By(0) = x0
and
Az(1) + By(1) = x1 .
(3.13)
8
CHAPTER 3. POPULATION GENETICS I
To complete the project of solving (3.12) it therefore remains to find two
independent solutions (z(t))t≥0 and (y(t))t≥0 . This is done by looking for
solutions of the form z(t) = rt . By substituting z(t) for x(t) in (3.12), we
find that r must satisfy, rt+1 = αr t + βrt−1 , or, dividing through by rt−1
r2 − αr − β = 0.
(3.14)
This is called the characteristic equation of the (3.12). If r is a root, then rt
is indeed a solution to (3.12). There are two cases to consider, according to
whether the characteristic equation has two distinct roots or only one root.
Case (i): If there are two distinct roots r1 and r2 , then we obtain two
distinct solutions r1t and r2t . Hence, by the superposition principle, x(t) =
Ar1t + Br2t is the general solution to (3.12).
Example 3.2.3. Solve
x(t+1) = −3x(t) − 2x(t−1),
x(0) = 1, x(1) = 0.
(3.15)
The characteristic equation is r2 + 3r + 2 = 0, which has roots r1 = −1 and
r2 = −2. Thus the general solution to (3.15) is x(t) = A(−1)t + B(−2)t .
The initial conditions require 1 = x(0) = A + B and 0 = x(1) = −A − 2B;
these are easily solved to find A = 2, B = −1. Hence x(t) = 2(−1)t − (−2)t .
Case (ii): Suppose the characteristic equation has a single root r. This
occurs when β = −α2 /4, and then the root is r = α/2. In this case, we
obtain immediately only the solution (α/2)t . However, it can be checked
that in this case t(α/2)t is again a solution; this is left as an exercise. Then,
the general solution has the form x(t) = A(α/2)t + Bt(α/2)t .
We turn now to the inhomogeneous equation, namely (3.11) with β 6= 0.
(This will not be needed later, but we include it for completeness.) There is
a variant of the superposition principle for the inhomogeneous equation: if
xp (t) is any particular solution to (3.11), then the general solution to (3.11)
can be written in the form
x(t) = Az(t) + By(t) + xp (t),
t ≥ 0,
where z(t) and and y(t) are independent solutions to the homogeneous equation (3.12), and A and B are arbitrary constants. This principle is again a
consequence of the linearity of the difference equation. To prove it, let (xp (t))
be a fixed particular solution to the inhomogeneous equation, and let (x(t))
be any other solution. We want to show x(t) = Az(t) + By(t) + xp (t). But
x(t+1) − xp (t+1) = [αx(t) + γx(t−1) + β] − [αxp (t) + γxp (t−1) + β]
= α (x(t) − xp (t)) + γ (x(t−1) − xp (t−1)) ,
3.1. FINITE DIFFERENCE EQUATIONS
9
since the β terms cancel out. This says that x(t) − xp (t) solves the homogeneous equation; hence, x(t) − xp (t) = Az(t) + By(t), and so x(t) =
Az(t) + By(t) + xp (t), as we wanted to show.
Therefore, to solve inhomogeneous equations we only need to come up
with some solution of the inhomogeneous equation. The easiest method is
to guess simple forms. For example the constant sequence, x(t) = x̄ for all
t ≥ 0 solves (3.11) if x̄ = αx̄ + γ x̄ + β. As long as α + γ 6= 1 this has
the unique solution x̄ = β/(1−α−γ), which therefore provides a particular
solution. The problem of finding particular solutions when α + γ = 1 is left
to the exercises.
Example 3.2.3, continued. Solve
x(t+1) = −3x(t) − 2x(t−1) + 1,
x(0) = 1, x(1) = 0.
(3.16)
We look for a constant solution of the form x(t) = w, t ≥ 0. Substituting in
(3.16), requires w = −3w−2w+1, or w = 1/6. We found previously that the
general solution to the homogeneous equation x(t+1) = −3x(t) − 2x(t−1)
is x(t) = A(−1)t + B(−2)t . Therefore the general solution to (3.16) is
x(t) = A(−1)t + B(−2)t + 1/6. The initial conditions require 1 = x(0) =
A + B + 1/6 and 0 = x(1) = −A − 2B + 1/6, and solving gives A = 3/2,
B = −2/3. Thus, the solution to (3.9) is x(t) = (3/2)(−1)t − (2/3)(−2)t +
1/6. Some algebra shows that x(t) = (−1)t (1/6)[9 − 2t+2 ] + 1/6, and
for t ≥ 2 this will oscillate between positive and negative values with ever
increasing amplitude, as t increases. Hence, the solution will not converge
to the constant solution 1/6.
The methods developed in this section can be extended to linear difference equations of any order.
3.1.3
Problems
Exercise 3.1.1. a) Calculate and plot the values x(0), x(1), x(2), x(3) for the
solution to
(i) x(t+1) = (1/2)x(t) + 2, x(0) = 2.
(ii) x(t+1) = 2x(t) + 2, x(0) = 2.
(iii) x(t+1) = (2/3)x(t) + (1/x2 (t)), x(0) = 3.
b) For case (i) of part a), identify the fixed point γ and observe that it is
stable. Determine how large n must be in order that |x(n) − γ| ≤ 0.01.
c) What is the fixed point of (iii) in part a)? Does the solution appear to
be converging to the fixed point?
10
CHAPTER 3. POPULATION GENETICS I
Exercise 3.1.2. Consider x(t + 1) = αx(t) + βx(t − 1), where 4β = −α2 ,
so that the characteristic equation has only one root r = −α/2. Show that
Art + Btrt solves the equation.
Exercise 3.1.3. The homogeneous, linear second order equation x(t+1) =
x(t) + x(t − 1), with initial conditions x(0) = 1 and x(1) = 1 defines the
Fibonacci sequence. Solve the equation to find a formula for x(t), for any
t ≥ 0.
Exercise 3.1.4. a) The object of this part is to find a particular solution to
x(t+1) = αx(t) + βx(t−1) + γ, when γ 6= 0, and α + β = 1. It does not have
a constant particular solution.
However, show that if, in addition, α 6= 2, there is a constant A such
that x(t) = At is a particular solution.
b) Show that if α = 2 and β = −1 there is a particular solution of the form
Bt2 .
c) Solve x(t+1) = (1/3)x(t) + (2/3)x(t−1) + 1, x(0) = 1, x(1) = 0.
Exercise 3.1.5. Find the solution of x(t+1) = −(5/6)x(t) − (1/6)x(t−1) + 1,
x(0) = 0, x(1) = 0. Show that the solution tends to the constant solution.
Exercise 3.1.7. Verify that the linear difference equation in Example 3.1.2
is Newton’s method for finding the root of x2 − 2. Calculate the next term
x(3) of
√ the solution
√ with initial condition x(0) = 1 and observe how close it
is to 2. In fact 2 is a stable fixed point for the difference equation and
x(0) = 1 is in its basin of attraction.
Exercise 3.1.7. (a) Consider the equation x(t+1) = αx(t) + g(t+1), where
(g(t))t≥1 is a given sequence. For convenience, define g(0) = 0. Show that
t
x(t) = Aα +
t
X
αt−s g(s) is a solution for any constant A.
s=0
(b) Consider x(t+1) = h(t+1)x(t) + g(t+1), the fully time-inhomogeneous,
first-order, linear difference equation . As before, set g(0) = 0. For 0 ≤ s < t,
define T (s, t) = h(s+1)h(s+2) · · · h(t); for all t ≥ 0, define T (t, t) = 1. Show
that
x(t) = AT (0, t) +
t
X
s=0
is a solution for any constant A.
T (s, t)g(s)
3.2. MODELING PRINCIPLES
3.2
11
Modeling principles for population genetics
In this section we develop the essential concepts required for building population genetics models. We discuss sex, genotype and generational structure,
define genotype and allele frequencies and their relationship to each other,
and present the basic mathematics of random mating. Throughout, the
study of one locus with two alleles serves as a running example.
3.2.1
Some biological considerations
Sex. (And you never thought you’d see this word in a math text!) Species
engaging in sexual reproduction are either monecious or dioecious. Each
individual of a monecious species houses both male and female sex organs,
so individuals are themselves neither male or female. (The root meaning
of “monecious” is in one house.) Plants with flowers that both contain an
ovum and produce pollen are examples of monecious species. In contrast,
individuals of dioecious (in two houses) species are either male or female.
The distinction between monecious and dioecious species is relevant to geneology, because any individual of a monecious species can mate with any
other individual. Potentially, a monecious individual can even mate with
itself, which is called self-fertilization or selfing. In reality, selfing is prevented in many monecious species by various biological mechanisms, which
are presumably adaptations preventing excessive inbreeding.
Autosomal vs. sex chromosomes. Genetic mechanisms of sex determination in dioecious species are diverse and complicate analysis of inheritance.
Sex is usually determined by a particular chromosome, the sex chromosome,
which is present in one sex, but not the other, or is present in different numbers. Chromosomes other than sex chromosomes are called autosomal. In
diploid species, autosomal chromosomes come in homologous pairs in both
male and female, so male and female genotypes for loci on autosomal chromosome have the same form. However, sex chromosomes may not be paired
and there may be but one locus per individual for the genes they carry, so
male female genotypes inclucing a locus on the sex chromosome will differ
in form. Genes or loci on sex chromosomes are said to be sex-linked.
We shall only study models for dioecious species following the human
pattern of sex determination. A human female has two paired X chromosomes, but a male has only one X, which is paired instead with a Y
chromosome. Genes on the X chromosome do not have loci on the Y chromosome, and so the male will carry only a single allele for these genes. He
will not pass genes on the X chromosome to his male offspring, because his
male offspring are precisely those that receive his Y chromosome, not his X
chromosome.
12
CHAPTER 3. POPULATION GENETICS I
3.2.2
Genotypes and Populations
Population genetics is mainly a study of genotype and allele frequencies of
populations. The concepts of genotype and allele are covered in Chapter
1. Genotypes are always defined with respect to some fixed set of loci that
we are interested in studying. However, we usually speak of the genotype,
assuming the loci of interest are fixed before hand and understood. The
genotype of an individual is a list of all the alleles it carries at the loci of
interest in the chromosomes of (any) one of its cells. Single letters are always
used to denote alleles, and hence genotypes always take the form of strings
of letters. For example, the pea plants of Mendel discussed in Chapter 1
admit two alleles for pea color: Y , for yellow, and G, for green. The peas
are diploid and hence there are two loci for color on the chromosomes of a
typical cell. The possible genotypes with respect to the locus for pea color
are thus Y Y , Y G, and GG. There are also two alleles, W (wrinkled) and S
(smooth) for pea texture, and again two loci for texture in the chromosome.
Hence the possible genotypes with respect to the loci for color and texture are
are Y Y SS, Y Y W W , Y Y W S, Y GSS, Y GW W , Y GW S, GGSS, GGW W ,
GGW S.
A population is a collection of living organisms. But population geneticists are interested only in genotypes, and so they suppress the extraneous
flesh and blood reality of individuals and treat a population merely as a
collection of genotype letter strings, one string per individual. For example,
to the population geneticist, {Y Y, Y G, GG, GG, Y Y, Y G, Y G} represents a
population of 7 pea plants in a study of the genetics of pea color. Or,
for another example, if you were to participate in a study of the genetics
of eye color, you would enter the data set as letters coding your eye color
alleles, and all your other distinctive features —your good looks, superior
intelligence, and athletic prowess—would be ignored. Sorry. Thinking of a
population as a collection of letter strings is very helpful to a clear understanding of the models.
3.2.3
Gene and Allele Frequencies
Consider a population and one of its possible genotypes, G1 · · · Gk . The
frequency fG1 ···Gk of G1 · · · Gk is simply:
4
fG1 ···Gk =
number of individuals with genotype G1 · · · Gk
.
population size
(3.17)
When time is a consideration, fG1 ···Gk (t) will denote a genotype frequency
at time t.
3.2. MODELING PRINCIPLES
13
Example 3.2.1. Consider the following population of a diploid species, in a
study of a single locus that admits two alleles A and a:
AA, AA, AA, Aa, Aa, Aa, Aa, aa, aa, aa, aa, aa
There are 12 individuals represented in the population, three of which are
AA and five are aa. Thus fAA = 3/12 = 1/4, and faa = 5/12.
Allele frequencies are computed by counting alleles only and ignoring how
they are organized into genotypes. Let ` be a locus. Given a population, the
allele pool for locus ` is the collection of all alleles from population genotypes
that occur at locus `. Recall that we think of a population as a collection
of genotype letter strings. Imagine collecting the letters denoting alleles at
locus ` from each individual and placing them in a box. That box then
contains the allele pool for `. Now consider a particular allele A that can
occur at `. The frequency fA of A is defined to be its frequency relative to
the allele pool corresponding to `:
4
fA =
number of A’s in the allele pool for `
size of the allele pool for `.
(3.18)
If ` is a locus on an autosomal chromosome of a diploid species, each individual genotype contains exactly two alleles at `. Therefore, if there are
N individuals in the population, the size of the allele pool equals 2N , and
this will be the denominator to use in computing fA . Again, when time is
a consideration, fA (t) will denote an allele frequency at time t.
Example 3.2.1, continued and extended. In the single locus genotypes of
example 3.2.1, all letters denote alleles at the locus containing A. The
allele pool is thus the collection of the 24(= 2 × 12) letters comprising these
genotypes. There are 24 letters total of which 10 are A’s: thus fA = 10/24 =
5/12.
Consider instead the population
AABb, AAbb, AAbb, AaBB, AaBB, Aabb, Aabb, aaBb, aaBB, aaBB, aabb, aabb,
where B and b denote alleles at a locus different than that of A and a. The
calculation of fA is unchanged, because the allele pool for ` is unchanged,
the numbers of B and b alleles being irrelevant. In calculating fA , one should
not divide by the total number, 48, of letters appearing in the population.
The reader should confirm that fb = 14/24.
Exercise 3.2.1. Consider a locus ` on the X chromosome in humans. The
gene has three alleles A, a and ā. Consider a second locus on an autosomal
chromosome admitting alleles B and b. Find fA and fB for a population
14
CHAPTER 3. POPULATION GENETICS I
consisting of 5 males and 7 females, where the males have genotypes ABB,
ABB, aBb, āBb, and ābb, and the females have genotypes AAbb, aaBb,
aābb, āABB, āāBB, aABb, and aāBB.
Genotype and allele frequencies cannot be arbitrary, but must obey simple algebraic constraints. We will illustrate this for a simple case; the student
should then be able to derive analogous relationships for other situations—
see, for instance, exercises 3.3.4 and 3.3.5.
Genotype and allele frequencies for the one locus/two allele case in a
diploid population. Let the alleles be denoted A and a; the possible genotypes
are then AA, Aa, and aa. Since each allele is either A or a, it follows that
fA + fa = 1.
Likewise,
fAA + fAa + faa = 1.
(3.19)
In addition allele and genotype frequencies are related as follows:
fA = fAA +
fAa
,
2
fa = faa +
fAa
.
2
(3.20)
We derive the first equation of (3.20), the second being similar. Let N denote the size of the population. Since the population is diploid, the size of
the allele pool for the locus at which A occurs is 2N . Now count the number
of A’s in the allele pool for ` in terms of the genotype frequencies. By definition of fAA , there are N fAA genotypes AA in the population, and so they
contribute a total of 2N fAA letter A’s. Similarly, there are N fAa genotypes
Aa contributing a total of N fAa letter A’s. Individuals of genotype faa of
course contribute no A’s. Hence, using definition (3.18),
fA =
2N fAA + N fAa
fAa
= fAA +
2N
2
In general, genotype frequencies cannot be recovered from allele frequencies, because they depend not just on numbers of alleles, but on how the
alleles are distributed among individuals.
3.2.4
Random Mating
The issue at the heart of modeling the evolution of genotype frequencies is:
• For each genotype G, what is the probability that an offspring of a
mating has genotype G?
3.2. MODELING PRINCIPLES
15
The purpose of this section is to define a particular model called random mating and deduce from it principles for calculating offspring genotype
probabilities. The concept of random sampling, as defined and discussed in
section 2.1, will play a central role. Recall that to “sample a population
at random” means to select an individual from it randomly so that each
individual has an equally likely chance of being chosen.
The idea of the random mating model is that each individual in the
mating pool has the same chance of reproductive success and that mate
choice is totally by random. Suppose this is true, and do the following
experiment. Sample an individual at random from the population of all offspring of the mating pool and record its parents. Then each possible pair
of parents should be equally likely as an outcome. But from section 2.1, we
know that this result is probabilistically the same as sampling the parent
population at random. Thus we can imagine that each individual is created
by mating randomly sampled parents. Informally, we say that the child
chooses its parents by random sampling, although, strictly speaking, this
is nonsensical! Once the parents are chosen, a mating takes place through
the union of two gametes, one egg and one sperm, from the respective parents. Now, each parent contains a pool of gametes produced by meiosis,
as described in Chapter 1. We will assume further that the gamete from
each parent is chosen by random sampling. Because the gametes of a single
individual may carry different alleles, this assumption expresses a biological hypothesis, namely, that gamete genotype does not confer any selective
advantage. Let us formalize this discussion by stating a precise definition.
Definition. A random mating is the creation of a new individual by
uniting the gametes (one egg and one sperm) of two parents chosen by
independent random samplings from the mating pool. Furthermore, the
gamete a chosen parent contributes is selected by a random sample of its
gametes.
The models we shall study all assume diploidy, and, in the case of dioecious species, the human model of sex determination. In these cases, random
mating leads to a simple principle for calculating offspring genotype probabilities. Let A be an allele and consider a random mating. For convenience,
let p1A denote the probability the first parent chosen in a random mating
passes allele A to the child and let fA1 be the frequency of A in the pool
from which this parent is drawn. Similarly define p2A and fA2 for the second
parent. Of course when the organism is monecious and both parents are
drawn from the same pool, then fA1 = fA2 .
16
CHAPTER 3. POPULATION GENETICS I
Lemma 1 In a random mating, the allele a child gets from each parent
at locus ` is a random sample from that parent’s allele pool for locus `.
Therefore,
piA = fAi i = 1, 2.
(3.21)
Proof: Fix a diploid locus `. If the genotype of an individual at locus ` is
A1 A2 ( A1 = A2 being possible), half its gametes contain a copy of A1 and
the other half copies of A2 . Hence, if it participates in a random mating,
A1 and A2 have the same probability (1/2) of being passed to its offspring
when it mates.
Consider now a random mating, choosing parent i from a population i.
Each individual of population i has an equal chance of being drawn, and, as
we have just seen, each of its alleles at locus ` has an equal chance of being
passed to the child. Therefore, in a random mating each allele in the allele
pool for locus ` of parent i has an equal chance of being passed to the child.
But this is equivalent to choosing the allele from the allele pool of parent i
by random sampling, which is what the lemma claims. Now we know from
(2.5) in Chapter 2 that, in a random sampling, the probability of drawing
type A is just the frequency of type A, and hence (3.21) follows.
A similar argument works for a locus on the Y chromosome.
This lemma contains just about everything we need to know in developing models based on random mating. We could almost take equation (3.21)
as a definition of random mating. But we have tried to lead up to it in
a careful way, to show how it is a consequence of the idea of completely
random mate choice and of the nature of sexual reproduction.
The next example shows how to compute genotype probabilities of offspring of random mating in the simplest situation.
Example 3.2.2. Random mating in the one locus/two allele case in a monecious population.
Consider a population of genotypes AA, Aa, and aa in a monecious
population. Then both parents are drawn from a common population and
so fA1 = fA2 = fA and fa1 = fa2 = fa . The probability that the offspring of
a random mating is AA is the probability that it gets allele A from each
parent, and since parents are chosen independently,
2
P (offspring is AA) = fA
= (fAA +
fA a 2
) .
2
(3.22)
Similarly,
fA a 2
) .
(3.23)
2
The event that a random mating produces an Aa is the union of the event
that the first parent contributes A and the second a, which has probability
P (offspring is aa) = fa2 = (faa +
3.2. MODELING PRINCIPLES
17
fA fa with the event that the first parent contributes a and the second A,
again having probability fA fa . Hence,
fA a
fA a
)(faa +
) (= 2fA (1 − fA ))
2
2
(3.24)
(Show that these three probabilities add up to 1!)
P (offspring is Aa) = 2fA fa = 2(fAA +
Remark. The definition of random mating stipulates that the parents are
sampled independently. If the species is monecious, both samples are from
the same pool of potential parents, and hence independent sampling can
lead to the same individual being chosen twice, which amounts to selfing.
However, when the population has size N , the probability of selfing is 1/N
(see exercise 3.2.4). For N of even moderate size, the selfing probability is
thus small, and then the independence assumption is a good approximation
even when selfing cannot occur.
Discussion.
The technical definition of random mating is abstract and far removed
from what mating means biologically in real populations. The following
remarks are meant to help the reader think more deeply about the modeling
issues arising here. The reader may pass over them for now if he or she
wishes, since they are not used in the sequel. However, we encourage reading
them eventually.
1. The term “random mating” as used here has a technically precise
sense. It refers not to any situation in which there is some random factors
in mating, but to the specific model in which mating is completely random,
in the sense that there are no mating preferences and no differences between
individuals in their chances of passing genes to offspring.
2. While the random mating model may look simple in hindsight, the
strategy used to derive it is subtle. A naive approach would attempt to base
a model directly on the actual physical process of mating. But this would require building probabilistic models for how many times an individual mates,
the number of offspring per mating, etcetera, leading to models which are
complex and species dependent. By taking a conditional viewpoint, looking
at the parents of a random individual, and just applying the idea that all
potential parents have equal chances, one is able to bypass the ultimately
irrelevant details of direct models.
3. The definition of random mating applies to the production of one offspring. We did not impose assumptions about how different random matings
from the same population are related probabilistically, for example, whether
they are independent or not. We return to this question later.
18
CHAPTER 3. POPULATION GENETICS I
4. What good is the random mating model? Surely there is an element
of randomness in all mating, but, especially with mammals, fish and birds,
individuals in general exercise mating preferences. Accurate models would
account for sexual preference by assigning higher probabilities to some pairings and lower probabilities to others. Is a model with completely random
mating ever useful?
There are several responses to this question. First, there are many situations, for example, pollination of plants by wind or insect, in which the
opportunity for sexual selection is limited. Then the random mating model
may indeed be a good approximation of reality. Second, random mating
may well describe gene mixing at selected loci, even if individuals themselves are not mating completely at random. For instance, consider blood
type, which is inherited genetically, and human mating. If you are looking
for a mate, surely you are exercising preferences about looks, personality,
earning potential, whether the person in question is a mathematician or biologist, etc. But probably you are not saying to yourself, if you live outside
of Transylvania,“I really go for those A blood types.” At least I hope not.
Therefore, if blood types are distributed evenly across all the traits that
do affect mate selection, mating will involve no preference of genotypes for
blood type, and random mating is then reasonable. Finally, random mating
is not proposed as a universal model. It is a baseline case whose virtue is
simplicity. In modeling one wants to start with the simplest assumptions
and explore what they imply before adding on further layers of complexity,
and models with random mating can be analyzed very completely. We can
gain insight into more complicated models by comparing them to the random mating model, and we can infer in nature when random mating is not
occuring by comparing data to the predictions of random mating models.
3.2.5
The infinite population assumption.
The models in this chapter are derived by considering what happens in the
limit as population size tends to infinity. This is called the infinite population
assumption, and the idea is that it should yield accurate models for large
populations. In mathematical practice, it means the following. Consider an
offspring population built by random mating from a given parent population.
Let fG1 ···Gk denote the frequency of a genotype in the offspring population,
and let pG1 ···Gk denote the probability that genotype is produced in a single
random mating. The infinite population assumption consists in imposing
for every genotype the identity
fG1 ···Gk = pG1 ···Gk .
(3.25)
To understand what this has to do with large populations, imagine creating
(N )
the offspring population by succesive random matings and let fG1 ···Gk be
3.2. MODELING PRINCIPLES
19
the genotype frequency in the first N offspring produced; it is the same
as the empirical frequency of G1 · · · Gk in the first N matings, and it is a
random variable. If the random matings were independent of each other,
(N )
then the law of large numbers would imply that limN →∞ fG1 ···Gk = pG1 ···Gk
with probability one. The definition of random mating did not actually
impose conditions on the correlation of different random matings. Still, it
is reasonable to expect that, as N grows, the dependence between matings
declines fast enough that the law of large numbers holds. Thus, identity
(3.25) represents what should happen, at least heuristically, in the limit
as population size tends to infinity. In practice, the infinite population
assumption means supposing the population under study is so large that
(3.25), which really ought to be an approximation, may be treated as an
exact identity. (A mathematically rigorous treatment would derive a law
of large numbers from models with specified correlations between matings.
However, in all elementary treatments, fG1 ···Gk = pG1 ···Gk is just assumed as
reasonable.)
The great, simplifying virtue of the infinite population assumption is
that it identifies genotype frequencies, which in real, finite populations are
random, with the determinstic quantities pG1 ···Gk . Hence, in infinite population models, genotype frequencies evolve deterministically.
3.2.6
Interaction of Generations
How individuals enter and leave the mating pool over time is an important
factor in gene flow. The simplest model assumes nonoverlapping generations. This means that the individuals of generation t mate among
themselves to produce generation t+1, and once this is done, mate no more;
generation t+1 mates to produce generation 2 and then mates no more, and
so on. This is a good model for species with annual reproduction cycles and
seasonal mating.
The extreme opposite of nonoverlapping generations is a model in which
births and deaths take place continually, and, as soon as an individual is
born, it enters the mating pool. In such a case, distinct generations are not
even well defined. Intermediate models postulate a generational structure,
but allow mating pools of different generations to mix in a limited way.
3.2.7
Problems
Exercise 3.2.2. (One gene/two alleles) Allele frequencies fA and fa = 1 − fA
do not determine the genotype frequencies fAA , fAa , and faa . Give two
different sets of genotype frequencies for which fA = 0.5.
Exercise 3.2.3. One locus/two allele case in a dioecious population. Consider
a locus on an autosomal chromosome in a population of a dioecious species.
20
CHAPTER 3. POPULATION GENETICS I
Let A and a denote the alleles that appear at the locus. Both males and
females have two copies of each locus, and so both males and females can
m , f m , etc.,
have each of the three possible genotypes AA, Aa, and aa. Let fAA
Aa
f
f
denote frequencies in the male subpopulation, and fAA , fAa , etc., frequencies
in the female subpopulation. A random mating is formed by choosing the
father by random selection from the male subpopulation and a mother by
random selection from the female subpopulation.
(i) Find two expressions for P(offspring is Aa), one in terms of allele
frequencies and the other in terms of genotype frequencies of both
male and female subpopulations.
(ii) Derive similar expressions for P(offspring is AA) and P(offspring is aa).
(iii) Let fA be the frequency of allele A in the entire population. In general,
m , f f , etc. However, find an
one cannot express fA in terms of fAA
AA
expression for fA in terms of these genotype frequencies and the ratio,
ρ = N m /N f , of the size of the male population to the size of the
female population.
Exercise 3.2.4. Calculate the probability that selfing occurs in a random
mating in a monecious population of size N .
Exercise 3.2.5. Probabilities in Mendel’s experiments For this problem it
may be helpful to refer to Chapter 1, Section 1. Consider the genotype for
pea shape and pea color in Mendel’s peas. For color, there are two alleles
Y and G for yellow and green and green is dominant. The two alleles for
shape are W for wrinkled and S for smooth and smooth is dominant.
(i) There are four possible phenotypes relative to these two traits: smooth,
green peas; smooth yellow peas; yellow, smooth peas; and yellow, wrinkled
peas. List the genotypes that give rise to each phenotype. You should have
9 genotypes in all.
(ii) Consider a plant with genotype Y GSW . The genotypes of the gametes of this plant will have one allele for color and one for shape: they will
be GS, GW , Y S, Y W . Assuming that the genes for color and shape assort
independently, show that the gamete genotypes are equally probable.
(iii) A random cross of Y GSW with itself consists of of a random sample
of size 2 from the gamete pool of Y GSW , one to choose the sperm and the
other the egg. The joining of their genotypes is the result of the cross. Determine the probability of each possible different phenotype that can result
from the cross.
Exercise 3.2.6. This problem uses Tchebysheff’s inequality and the Central
Limit approximation of the binomial distribution; see Chapter 2, Sections
3.3. MODELS WITH NO SELECTION
21
2.3.6 and 2.3.7. Consider the one locus/two allele case. Let the frequency
of allele A in a parent population be fA = 2/3. Assume that the first generation contains N individuals produced by N independent random matings.
Define fAA as in Section 3.2.6. This problem shows how to get an idea the
probabilities of deviation of fAA (1) from its mean 4/9 for a population of
size 1000.
(i) If N = 1000, use Tchebysheff’s inequality to find an upper bound on
the probability that fAA (1) differs from fA2 = 4/9 by more than 0.05.
(ii) If N = 1000, use the DeMoivre-Laplace Central Limit Theorem to
estimate the probability that fAA (1) differs from fA2 = 4/9 by more than
0.05. Note that this approximation gives a better result than the Tchebysheff
inequality, which in the binomial case is not sharp.
3.3
Models with no selection
This section studies models for the evolution of genotype frequency when
no selection is acting, primarily for the case of one locus with two alleles
in a diploid species. We shall start with the simplest model and gradually
complexify by modifying the basic assumptions.
3.3.1
Basic model
The basic model studies the case of one locus with two alleles under the
following assumptions:
Random mating.
(A.1)
Nonoverlapping generations.
(A.2)
Infinite population.
(A.3)
Monecious species.
(A.4)
No selection, mutation, or migration.
(A.5)
The meaning of these assumptions, except for the last one, was explained
in the previous section. Mutation refers to a random change in an allele
of a parental gamete due to physical changes in coding DNA, caused for
instance by copying error in meiosis or by radiation damage. Mutation
alters the basic rule stated in Lemma 2.1 for the probability of transmission
of an allele. Migration adds genotypes from outside sources to a population.
Selection occurs when different genotypes have different effects on survival
or reproductive success. Thus, when selection acts, the genotype frequencies
within one generation can change over time, as that generation matures and
the less fit individuals die off, or random mating will no longer be valid,
22
CHAPTER 3. POPULATION GENETICS I
since there will be individual differences in mating success. These processes
are all excluded in the basic model.
Consider a locus with two alleles A and a. Assumptions (A.1)–(A.5)
lead directly to a mathematical model for the evolution of the associated
genotype frequencies. In this model, fAA (t), fAa (t), faa (t), fA (t), and fa (t)
will denote the genotype and allele frequencies of generation t, t ≥ 0. These
are unambiguously defined, first because assumption (A.2) means that each
generation is a coherent entity, and second, because assumption (A.5) implies that the frequencies in each generation remain constant from the time
of birth to the time of reproduction. The model itself will consist of a set
of first order difference equations for the genotype frequencies, derived as
a logical consequence of the assumptions. The derivation is in fact simple
using the work we have already done in the previous section. Consider, for
example, fAA (t + 1), for any t ≥ 0. According to (A.2), generation t + 1
is produced by random matings of parents in generation t, and fAA (t+1)
equals the frequency of AA in the offspring born to generation t parents.
Therefore, by the infinite population assumption, (3.25),
fAA (t+1) = P (a random mating of generation t parents produces AA) .
But we saw in Example 3.2.2, equation (3.22), how to calculate this probability: it is fA2 (t). Also from (3.20), we know fA (t) = fAA (t) + (1/2)fAa (t).
Thus,
fAa (t) 2
2
.
(3.26)
fAA (t+1) = fA (t) = fAA (t) +
2
The same reasoning yields as well,
fAa (t+1) = 2fA (t)fa (t) = 2 fAA (t)+
faa (t+1) =
fa2 (t)
fAa (t)
= faa (t) +
2
fAa (t)
2
faa (t)+
fAa (t)
(3.27)
2
2
(3.28)
These three difference equations, (3.26), (3.27), and (3.28), constitute the
basic model.
At first glance, the equations (3.26), (3.27), (3.28) look too complicated
to solve explicitly, since they consist of three coupled, nonlinear difference
equations. But they hide an underlying simplicity that reveals itself if we
ask how the allele frequencies evolve. Using first (3.20) and then (3.26),
(3.27), and fA (t) + fa (t) = 1,we find,
fAa
(t+1)
2
= fA2 (t) + fA (t)fa (t) = fA (t)(fA (t) + fa (t))
fA (t+1) = fAA (t+1) +
= fA (t).
(3.29)
3.3. MODELS WITH NO SELECTION
23
Thus, the model reduces to the trivially simple equation
fA (t+1) = fA (t).
It solutions are constant in time: fA (t) = fA (t − 1) = · · · = fA (0) for
all t ≥ 0. Under the basic model, allele frequencies do not change from
generation to generation! And since allele frequencies determine genotype
frequencies by equation (3.20), we obtain the complete solution to (3.26),
(3.27), and (3.28): for all t ≥ 1 the genotype frequencies take on the constant
values
fA (t) = fA (0) = fAA (0) + (1/2)fAa (0),
2
fAA (t) = fA (t−1) =
fa (t) = 1 − fA (t) = 1 − fA (0)
fA2 (0);
fAa (t) = 2fA ((t−1)fa (t−1) = 2fA (0) (1 − fA (0))
faa (t) =
fa2 (t−1)
(3.30)
2
= (1 − fA (0)) .
This solution is so important that it is given a special name, in honor of
the first researchers to state it clearly.
Definition: Allele frequencies fAA , fAa , and faa , with fAA + fAa + faa = 1,
are said to be in Hardy-Weinberg equilibrium if there exists a p, 0 ≤ p ≤ 1,
such that
fAA = p2 ,
fAa = 2p(1 − p),
faa = (1 − p)2 .
Using this definition, we can summarize the results of our analysis so far
as follows.
Theorem 1 (Hardy-Weinberg Theorem) Assume (A.1)–(A.5). Then
the allele frequencies are constant, and, for all generations t ≥ 1, the genotype frequencies for AA, Aa, and aa are in Hardy-Weinberg equilibrium with
p = fA (0) = fAA (0) + fAa (0)/2.
Although simple, this is an extremely important result. Biologically,
it says that in the absence of selection, random mating maintains
genetic variation in the infinite population model. And it specifies
the genetic equilibrium quantitatively. In a natural population, absence of
Hardy-Weinberg equilibrium indicates that one of the assumptions (A.1)–
(A.5) does not hold. If the population is large and isolated and random
mating seems likely, it is reasonable to deduce that selective pressure or
mutation is acting to maintain disequilibrium.
Testing for Hardy-Weinberg equilibrium is simple, due to the following
criterion, which you are asked to derive in Exercise 3.3.2.
24
CHAPTER 3. POPULATION GENETICS I
Genotype frequencies fAA , fAa and faa are in Hardy-Weinberg
equilibrium if and only if
2
fAa
= 4fAA faa .
(3.31)
Example 3.3.1. Assume (A.1)–(A.5), and let fAA (0) = 0.2, fAa (0) = 0.4,
and faa (0) = 0.4. Describe the evolution of genotype frequencies.
In generation 0, the population is not in Hardy-Weinberg equilibrium,
because (fAa /2)2 (0) = (0.2)2 = .04 is not equal to fAA (0)faa (0) = (0.2)(0.4) =
0.08. The frequency of allele A is fA (0) = fAA (0) + (fAa /2)(0) = 0.4.
The Hardy-Weinberg theorem says genotype frequencies arrive at HardyWeinberg equilibrium with p = 0.4 in one generation. Thus, for t ≥ 1,
fAA (t) = (0.4)2 = 0.16, 2fAa (t) = 2(0.4)(0.6) = 0.48, and faa (t) = 0.36. The simplicity of Theorem 1 suggests it should have much simpler derivation than we have given, and indeed it does. Recall Lemma 1, which was
a crucial element in deriving the model (3.26), (3.27), (3.28). It says that
the probability pA (t) that a parent of generation t passes allele A to its
offspring in random mating is pA (t) = fA (t). In effect, the allele pool of
generation t + 1 for the locus of A is built by repeated random samplings
(two independent samples per mating) of the allele pool of generation t).
From the infinite population assumption, which is essentially a law of large
numbers, we should expect fA (t + 1) = pA (t). But then it follows that
fA (t+1) = pA (t) = fA (t), and we are done. In effect, the random matings
producing each generation completely and randomly redistribute allele pools
among the individuals of the next generation. This is why Hardy-Weinberg
equilibrium is achieved with the first generation.
Why didn’t we develop the Hardy-Weinberg theorem by this simple
method in the first place, other than our natural, professorial desire to confuse students?! First, from the purely logical standpoint, we developed the
infinite population assumption not for allele frequencies but for genotype
frequencies. The derivation in (3.29) that allele frequencies are constant
amounts to a proof that fA (t+1) = pA (t) from the infinite population assumption as stated in (3.25). Also, equations (3.26), (3.27), (3.28), while
complicated, follow more directly from viewing mating as a process producing new individuals rather than new allele pools. Finally, the student
will need to understand the analysis at the genotype level for dealing with
more complicated models, especially those with selection, in which genotype
affects reproductive success.
3.3.2
Problems
Exercise 3.3.1. You are studying a hypothetical species of butterfly. It has
one gene that controls wing color with two alleles, B and Y . Genotype
3.3. MODELS WITH NO SELECTION
25
BB butterflies have blue wings, genotype Y Y butterflies have yellow wings,
and genotype BY butterflies have green wings. You sample butterflies in a
population of mixed colors and find that the frequencies of blue, yellow and
green butterflies are, respectively, 0.2, 0.3 and 0.5. Is the population exactly
in Hardy-Weinberg equilibrium? If not, what would the Hardy-Weinberg
equilibrium be given the actual allele frequencies?
Exercise 3.3.2. a) Show that genotype frequencies fAA , fAa and faa are
2 = 4f
in Hardy-Weinberg equilibrium if and only if fAa
AA faa . (Remember,
fAA + fAa + faa = 1.)
b) The possible values of K = fAA and M = faa are definied by the
region K ≥ 0, M ≥ 0, and K + M ≤ 1. Graph this region in the (K, M )
plane. Derive a relation that expresses M = faa as a function of K = fAA
when they are in Hardy-Weinberg equilibrium and graph this curve in your
region.
2 −
c) For any valid set of frequencies fAA , fAa and faa , show that | fAa
4fAA faa | is bounded by 1. This shows that the difference is never too large,
so what may look like a small difference, may correspond to a situation far
from Hardy-Weinberg equilibrium.
Exercise 3.3.3. A large monecious population (size N ) of AA homozygotes is
brought into contact with a population of aa homozygotes of size 2N . From
that point on the populations merge at once and random mating takes place.
There is no selection, mutation or migration. Assuming N is large enough
that we may assume the infinite population hypothesis is valid, describe the
evolution of the gene and allele frequecies in all future generations.
3.3.3
The basic model for multiple alleles of one gene
This section is in the nature of a long exercise, We continue to study the
genotype for just one gene, but this time we assume that it admits m alleles,
labelled A1 , . . . , Am , where m ≥ 3. Therefore, the possible genotypes are
the pairs Ai Ai , where i and j range between 1 and m. For notational
convenience, denote the frequency of genotype Ai Aj in generation t by fij (t)
instead of fAi Aj (t). Similarly, let fi (t) be shorthand for the allele frequency
fAi (t).
The exercises that follow guide you toward a statement of the HardyWeinberg theorem for multiple alleles. They can be solved by straightforward generalization of the two allele analysis presented in the previous
section. If it helps, do them for the more concrete case m = 3 rather than
general m. This case contains all the ideas and shows why the general case
works.
Exercise 3.3.4. For the two allele case we know that fA = fAA + (fAa /2).
26
CHAPTER 3. POPULATION GENETICS I
Work out the analogous formula for the multi-allele case. That is, for each
i, express fi (t) in terms of fij (t), 1 ≤ j ≤ m.
Exercise 3.3.5. Apply random mating to express fij (t+1) in terms of the
allele frequencies fA1 (t), . . . , fAm (t) in the previous generation.
Exercise 3.3.6. Generalize the Hardy-Weinberg theorem to the multi-allele
case, as follows. Use the results of exercise 3.3.5 to show that allele frequencies are constant and that the genotype frequencies reach equilibrium values
in generation t = 1 and thereafter remain fixed. Express those equilibrium
genotype frequencies in terms of the allele frequencies in the population at
time t = 0, and define a generalization of Hardy-Weinberg equilibrium.
Rederive the Hardy-Weinberg equilibrium by arguing directly that allele
frequencies are constant.
Exercise 3.3.7. (See Exercise 3.3.2a).) Show that a set of genotype frequencies fij , 1 ≤ i ≤ j ≤ m, is in Hardy-Weinberg equilibrium, that is, will
remain constant for all future generations, if and only if for every i 6= j,
fij2 = 4fii fjj .
3.3.4
One gene/two alleles for dioecious populations
In this section we analyze what happens to the basic model when it is
assumed the species is dioecious rather than monecious. All the remaining
assumptions (A.1), (A.2), (A.3), and (A.5) are in force.
For dioecious species, loci on autosomal chromosomes must be analyzed
separately from loci on the sex chromosome, because these cases differ in how
sex and genotype are jointly inherited. Consider autosomal chromosomes
first. Recall from Chapter 1 Mendel’s hypothesis that different chromosomes
segregate independently of one another in meiosis. This means autosomal
chromosomes are inherited independently of the sex chromosome, and hence,
for genotypes with respect to loci on autosomal chromosomes,
sex and genotype are passed to progeny independently.
(A.6)
This principle will be adopted as another assumption in deriving the dioecious model. It has the following consequence. Let a1 · · · ak denote any
genotype with respect to loci on autosomal chromosome. Let pm
a1 ···ak (t+1)
denote the probability that a male offspring of a random mating of parents
of generation t produces the genotype a1 · · · ak , and let the pfa1 ···ak (t+1)
denote the probability of the same event for a female offspring. Then
f
pm
a1 ···ak (t+1) = pa1 ···ak (t+1)
(3.32)
Consider now a locus on an autosomal chromosome with an allele A.
The infinite population assumption, applied separately to male and female
3.3. MODELS WITH NO SELECTION
27
subpopulations, says that for all t ≥ 0,
m
fAA
(t+1) = pm
AA (t+1)
f
and fAA
(t+1) = pfAA (t+1),
f
where pm
AA (t+1) and pAA (t+1) are as defined in the previous paragraph. It
m (t+1) = f f (t+1) for all t ≥ 0. This
immediately follows from (3.32) that fAA
AA
reasoning applies to any genotype. Hence for the dioecious model, the the
male genotype and allele frequencies are equal to the correponding female genotype and allele frequencies in every generation t
starting with t = 1. Refering back to Exercise 3.2.2, do the following
problem to complete the analysis of the dioecious case.
Exercise 3.3.8. Consider a single locus with two alleles A and a. Assume
m (0), f m (0), f m (0) and f f (0), f f (0), f f (0) are given
the frequencies fAA
aa
aa
Aa
AA
Aa
and that assumptions (A.1), (A.2), (A.3), (A.5), and (A.6) are in force.
a) Calculate the genotype and allele frequencies, of the first generation,
in terms of the generation 0 genotype frequencies. Then calculate the allele
and genotype frequencies of the second generation.
b) Show that the allele frequencies of the male and female populations
are equal and constant for all generations t ≥ 1. Show that the genotype
frequencies are in Hardy-Weinberg equilibrium with
f
f
m
m
p = (1/2) fAA
(0) + (fAa
/2)(0) + fAA
(0) + (fAa
/2)(0)
in generations 2, 3, . . . .
Exercise 3.3.8 shows that autosomal gene frequencies for dioecious species
differ in only a minor way from the monecious case. The fact that there are
separate sexes only means that it takes two generations instead of one for
random mating to mix alleles sufficiently that Hardy-Weinberg equilibrium
is attained.
Next, consider a locus ` carrying alleles A and a on the X chromosome of
a human population. In this case (A.6) is no longer true. A female offspring
has two X chromosomes, one from the father and one from the mother,
and hence receives one allele from each parent. A male offspring receives a
Y chromosome from the father, which does not carry a locus for the gene.
Males only obtain an allele for ` from their mother. Hence sex and alleles
are not inherited independently.
Since each male carries only one X chromosome its genome will carry
only one allele for `, and so the male genotypes are the single letters A or a.
m (t), f m (t), f m (t) to contend with,
In this case there are no frequencies fAA
aa
Aa
m
m
m
only fA (t) and fa (t) = 1−fA (t). The females have the usual genotypes
AA, Aa and aa, whose frequencies are indicated as usual with a superscript
28
CHAPTER 3. POPULATION GENETICS I
f . Assume again that generations do not overlap, that there is no selection,
migration or mutation, and that mating is random. The next exercise guides
the reader through the formulation and analysis of the model under these
assumptions. It will turn out that the frequencies do not attain HardyWeinberg equilibrium in a finite number of steps, but do tend to HardyWeinberg equilibrium as time progresses. The reader will need to solve a
second-order linear difference equation to complete this problem and should
refer to Section 3.1 to learn how to do this.
Exercise 3.3.9. a) Show that the random mating and infinite population asf
sumptions imply fAm (1) = fAf (0), fAA (1)f = fAm (0)fAf (0), fAa
(1) = fAm (0)(1−
f
f
f
m
f
m
fA (0)) + fA (0)(1 − fA (0)), faa (1) = (1 − fA (0))(1 − fA (0)). Deduce that
fAf (1) = (fAm (0) + fAf (0))/2.
b) The same argument as in a) shows that for any t,
fAm (t+1) = fAf (t),
t ≥ 0;
1 m
(f (t) + fAf (t))
fAf (t+1) =
2 A
1 f
=
(f (t − 1) + fAf (t))
2 A
t≥0
t ≥ 1.
Isolating the first and last expressions of the second equation,
1
fAf (t+1) = (fAf (t − 1) + fAf (t)).
2
(3.33)
This is a second order, homogeneous linear difference equation. By solving
it—see section 3.1.2—find a formula for fAf (t) in terms of t, fAf (0), and
fAm (0).
f
f
f (t) at any
c) Express the genotype frequencies fAA
(t), fAa
(t), and faa
time t in terms of fAf (t) and fAf (t−1).
4
4
f
f
d) Find the limits fAf (∞) = limt→∞ fAf (t), fAA
(∞) = limt→∞ fAA
(t),
f
f
f
m
f
etc. in terms of fA (0) and fA (0). Show that fAA (∞), fAa (∞) and faa (∞)
are in Hardy-Weinberg equilibrium.
3.3.5
Infinite population with mutation, but no selection
In this section we study again a single locus admitting two alleles A and
a. We shall impose the assumptions (A.1)-(A.4) and shall also exclude
migration and selection. However, we want to allow mutation. Mutation
occurs when an allele in a parental gamete becomes modified in the course
of mating. Mutations, which can be induced by copying errors or exposure
to chemical toxins or radiation, cannot be predicted and so are considered
3.3. MODELS WITH NO SELECTION
29
to be random. One simple model supposes that A can mutate into a and
vice-versa according to the following rule:
(A.7); In each random mating and for each parent independently, an
allele A being transmitted to the offspring mutates to a with probability
u, and an allele a being transmitted to the offspring mutates to A with
probability v. Moreover, 0 < u + v.
Frankly, the motivation for this model is not really scientific. Rather, we
want to explore as an exercise how the basic model might change as a result
of mutation, and (A.7) is the simplest mutation mechanism one can think
of. It is what is called in the trade a “toy model.” The technical assumption,
0 < u + v, excludes the case in which u = v = 0 and no mutation occurs.
In deriving the basic model we worked first with genotype frequencies
and derived equations (3.26), (3.27), (3.28). But then (page 24) we gave a
simpler analysis using just allele frequencies. We adopt the latter approach
for the mutation problem. The infinite population assumption says that
fA (t+1) = probability an offspring
acquires A from a randomly selected parent
We need to compute the probability on the right-hand side. According to
(A.7), the offspring acquires A either if the parent contributes a gamete with
genotype A and A does not mutate, or if the parent contributes a gamete
with genotype a and a does mutate. The probability of the first event is
(1 − u)fA (t), since we know from Lemma 1 that the probability a randomly
selected parent of generation t contributes a gamete with genotype A is
fA (t) and the probability that A does not mutate is 1 − u. Similarly, the
probability of the second event is v(1 − fA (t)). Therefore,
fA (t+1) = (1 − u)fA (t) + v(1 − fA (t)) = v + (1 − u − v)fA (t).
(3.34)
This is a first order, linear difference equation. According to formula (3.5)
developed in section 3.2.1, its solution is
fA (t) = (1 − u − v)t [fA (0) −
v
v
]+
.
u+v
u+v
(3.35)
The next exercise asks the reader to show how this solution behaves in the
long-time limit.
Exercise 3.3.10. Verify this solution using (3.5). Assuming 0 < u + v < 2,
prove limt→∞ fA (t) = v/(u + v). Derive from this the following conclusions:
if v > 0 and u > 0, variation between A and a alleles is maintained; ;f v = 0,
while u > 0, allele A disappears in the long-time limit, while if u = 0 and
v > 0, allele a disappears.
Finally, determine limt→∞ fAA (t) and limt→∞ fAa (t).
30
CHAPTER 3. POPULATION GENETICS I
Exercise 3.3.11. Analyze the solution of (3.34) when u = 1 and v = 1 and
interpret.
(1)
Exercise 3.3.12. Let fA (0) = 0.5, Let fA (t) denote the allele frequency in
(2)
generation t in the the case in which u = 1/4 and v = 1/2. Let fA (t) be
the allele frequency in generation t in the case that u = 1/16 and v = 1/8.
(i) Show that the limiting frequency, as t → ∞, is the same in both
cases.
(ii) Denote the limiting frequency found in i) by p̄. Find the smallest
(1)
value T1 such that |fA (T1 ) − p̄| ≤ 0.01. Find the smallest value T2 such
(2)
(1)
(1)
that |fA (T2 ) − p̄| ≤ 0.01. (Note that fA (t) and fA (t) are both increasing
in t.) Compare T1 and T2 and explain why your result is to be expected on
intuitive grounds, considering the mutation rates in both cases.
3.3.6
A model with overlapping generations
In non-overlapping generation models, each generation produces the next
and then mates no more. One concrete way to picture this is to imagine
that mating takes place only during regularly spaced mating seasons, in
each of which the entire population mates to produce offspring and then
dies, thereby entirely replacing itself. Suppose instead that the population
replaces only a fraction h of itself in each mating season, as follows. The entire population first mates randomly to produce enough offspring to replace
a fraction h of itself. Once this is done, a fraction h of the parent population
(not the newborns) is eliminated by random selection and replaced by the
newborns. By the next mating season, the new arrivals are assumed to be
sexually mature and to participate on an equal footing in random mating.
This is a simple system in which mixing occurs between the genes of individuals born at different times. It is the purpose of this section to derive
the corresponding mathematical model when the other assumptions of the
basic model—(A.1), (A.3), (A.4), and (A.5)—are in force.
For the derivation, if h is the fraction of the population replaced, it is
convenient to measure time so that the mating seasons occur at times h,
2h, 3h, and so on. This is the proper time scale to compare models with
different values of h. In a time interval of length 1, approximately 1/h
mating seasons occur, in each of which a fraction h of the population is
replaced. Thus, independently of the value of h, the total number of new
individuals entering the population in one unit of time is equal to the size
of the population.
In what follows fA (h), fA (2h), · · · shall denote the allele frequencies in
the population at the end of each successive mating season. Thus fA (h)
is the allele frequency after the first birth, death and replacement occurs,
3.3. MODELS WITH NO SELECTION
31
fA (2h) the frequency after second occurence, and so on. Similarly, fAA (kh)
will represent the genotype frequency at the end of the k th mating season.
Consider genotype AA and let t and t + h denote the times of two successive mating seasons. The difference fAA (t + h) − fAA (t) will be a sum
of the changes due births of new individuals minus those due to deaths in
the mating season t + h. Now, the mating at time t + h occurs in a parent
population in which the frequence of allele A is, by definition, fA (t). The
random mating and infinite population assumptions thus imply that the
frequency of AA among the offspring is fA2 (t). The frequency of AA in the
parent population at time t + h is by definition fAA (t), and since death is
by random selection, the frequency of AA among the individuals selected
to die must be fAA (t) as well. Since a fraction h of the population is being
replaced, it follows that
h
i
fAA (t + h) − fAA (t) = h fA2 (t) − fAA (t)
(3.36)
(If you are not convinced of this, imagine that the population has size N and
compute the number of AA genotypes entering and leaving the population
in the mating season t + h.)
A similar analysis applies to fA (t + h). The allele frequency among the
offspring produced in the replacement event at t + h is fA (t). because we
know this frequency does not change in random mating. The proportion of
individuals selected for elimination with genotype AA is, as we know, fAA (t),
and the proportion with genotype Aa is, similarly, fAa (t). As a consequence,
the frequency of A among those eliminated is fAA (t)+fAa (t)/2 = fA (t), also.
This means that the number of A entering and leaving the population is the
same in each mating season and therefore the allele frequencies remains
constant over time, just as in the basic model.
Now apply the fact that allele frequency remains constant (so, fA (t) =
fA (0) for all t) to (3.36). The result is the following difference equation for
fAA (t):
fAA (t + h) = [1 − h]fAA (t) + hfA2 (0)
(3.37)
Except for the fact that the sequence fAA (0), fAA (h), fAA (2h), . . . is indexed
by multiples of h rather than by integers, this is a linear difference equation
of first order. By using Proposition 1, one can show that its solution is
h
i
fAA (kh) = (1−h)k fAA (0) − fA2 (0) + fA2 (0)
and that
4
fAA (∞) = lim fAA (kh) = fA2 (0).
k→∞
(3.38)
(3.39)
A similar analysis applied to the genotypes Aa and aa shows
4
fAa (∞) = lim fAa (kh) = 2fA (0)fa (0),
k→∞
4
faa (∞) = lim faa (kh) = fa2 (0).
k→∞
(3.40)
32
CHAPTER 3. POPULATION GENETICS I
Exercise 3.2.13. Verify (3.38)–(3.40) in detail, and show that the limiting
genotype frequencies are in Hardy-Weinberg equilibrium.
The equation (3.36) can be used to derive a continuous time model by
taking a limit as h ↓ 0. When h is very small, the population is mating
very frequently and replacing only a small fraction of itself each time. Thus
the limit as h ↓ 0 corresponds to a situation in which mating and death
occur continuously and at the same rate and in which offspring enter the
mating pool immediately upon birth. This is the extreme opposite of nonoverlapping generations. To obtain the continuous time model, divide both
sides of equation (3.36) by h and let h ↓ 0. The result is
0
fAA
(t) = lim
h↓0
fAA (t + h) − fAA (t)
= −fAA (t) + fA2 (0).
h
(3.41)
Exercise 3.3.14. Let fAA (0) = fAA . Solve equation (3.41) in terms of fAA ,
fA (0), and t. (Hint: consider f˜AA (t) = fAA (t)−fA2 (0), and find a differential
equation for f˜AA (t).) Show that limt→∞ fAA (t) = fA2 (0), which is the Hardy
Weinberg equilibrium value.
3.3.7
Conclusion
Hardy-Weinberg equilibrium emerges over and over in the models we have
derived. Indeed, one should expect it to occur, at least in the limit, in any
infinite population when random mating occurs and selection and mutation
do not operate, because random mating mixes the allele pool of each locus by
random selection. Thus, when the population is monecious and generations
do not overlap, random mating is able to completely mix the gene pool in
one mating season and Hardy-Weinberg equilibrium is attained in the first
generation. The other models of the chapter mainly deal with conditions
that limit the mixing random mating can do in each mating season. Thus,
if the population is dioecious, an extra generation is needed to mix the
male and female allele pools. If generations overlap, random mating can
only achieve partial mixing. But if in a model, extra mixing can occur
with each mating season, the population should tend to Hardy-Weinberg
equilibrium. The close relationship between random mating and HardyWeinberg equilibrium is the takehome lesson of this section.
3.3.8
Exercises
Exercise 3.3.15. Develop a one locus/two allele model for overlapping generations, as in Section 3.3, but with the following twist. Assume that mating
seasons occur at times h, 2h, . . . and in each season a fraction h of the population is replaced. However, assume that newborns require two seasons
3.3. MODELS WITH NO SELECTION
33
to mature, and do not mate in the first season after they are born, but
only in the second. However, newborns have the same probability as older
individual to be removed, so they may not survive until sexual maturity.
For modeling one needs at the start to keep track of the frequencies in the
mating and non-mating portions of the population. To initialize the system,
imagine at time zero that the entire is a population ready to mate in the
first season and let fAA (0), fAa (0), faa (0) fA (0) denote the frequencies in
this generation. If possible, find a solution to your model.
Exercise 3.3.16. Here is a model of one locus–two alleles in which mating is
not fully random. Assume otherwise an infinite population, non-overlapping
generations, and no selection.
Assume the population is composed of two subpopulations I and II, and
let A and a denote the alleles at the locus of study. Use f I (t) and f II (t) to
denote the frequency of allele A in each subpopulation for generation t. Now
assume the next generation is produced as follows. To replace an individual
in population I, determine its first allele by choosing an allele at random
from population I. Determine its second allele by drawing an allele at
random from population I with probability .8 and an allele at random from
population II with probability .2 (Note: .8 is the probability of choosing
population I, not the probability of drawing a specific allele!).
To replace an individual in population II, draw the first allele at random
from population II. Determine the second allele by drawing at random from
population I with probability .1 and from population II with probability
.9.
a) Assuming no selection, derive equations that determine f I (t + 1) and
+ 1) in terms of f I (t) and f II (t). Your equation should reduce to a
set of two linear update equations.
f II (t
b) Show using the result of a), that (f I (t)/2) + f II (t) is constant from
one generation to the next.
c) limt→∞ f I (t) and limt→∞ f II (t) both exist. Guess on the basis of
intuition what these limits are.
d) By using part b), find a single update equation for f I (t) and solve it.
Verify your guess for limt→∞ f I (t).
Exercise 3.3.17. (Two locus model.) Let the alleles at locus `1 be A1
and A2 . Let the alleles at locus `2 be B1 and B2 . Suppose that `1 and
`2 are on different chromosomes. By Mendel’s laws, these chromosomes
segregate independently. What is the evolution of the genotype frequencies
fAi Aj Bk Bm (t)? Describe the values of these frequencies explicitly in terms of
the allele frequencies of generation 0. Assume all the hypotheses (A.1)–(A.5)
of the basic model.
34
CHAPTER 3. POPULATION GENETICS I
Exercise 3.3.18. The problem with the blending hypothesis. In so far as
there was a theory of heredity in the late 1800’s prior to the rediscovery
of Mendel’s results, it would have been a theory of blending. That is, any
specific trait of an individual would be a blend half-way between the traits
of its parents. Ewens, in his text, Mathematical Population Genetics
(1979), says:
“It is easy enough to see that, with random mating, the variance in a
population for any characteristic would, under a blending theory, decrease
by a factor of one-half in each generation.”
This problem is about understanding Ewen’s statement. We need a
model. Suppose we have a characteristic whose strength can be described
by a continuum of values, x, say height. Let the random variable X denote
the height of an individual drawn at random from the population. Let us
interpret the “variance in a population for the height characteristic” as simply the variance of X. Denote this variance as σ 2 . Now what happens in
random mating? We draw and individual from the population at random
and let X1 denote its height. We draw a second individual from the population independently of the first and denote its height as X2 . The random
variable X1 and X2 are independent and have variance σ 2 . We mate the
two individuals to produce an offspring with height (1/2)(X1 + X2 ); this is
the blending. Now finish the reasoning!
Comment: Notice that under blending the variance of any characteristic over a population must disappear. (Why?) This is in total opposition
to the Hardy-Weinberg result, which shows that for inheritance of discrete
traits, variation is maintained. The blending theory is not consistent with
real populations, which do maintain variation.
3.4
An Infinite Population Model with Selection
We shall study a standard infinite population model with random mating
and selection. In this case, the difference equations determining the evolution of allele frequency are non-linear. In general, non-linear difference
equations do not have closed-form solutions. However there are some simple
tools for analyzing the qualitative behavior of solutions that will allow us to
understand the selection models. These tools are described first.
3.4.1
Nonlinear, first-order difference equations
There is a simple graphical technique, called cobwebbing, that allows one to
visualize solutions of first order difference equations of the form:
x(t + 1) = φ(x(t))
(3.42)
3.4. AN INFINITE POPULATION MODEL WITH SELECTION
35
It will always be assumed that the function φ is continuous. By itself,
cobwebbing is not a rigorous mathematical method. But it helps in guessing
the long term asymptotic behavior of solutions and in interpreting rigorous,
analytic techniques. In particular, cobwebbing makes clear heuristically how
solutions behave near fixed points. Recall from section 3.1 that x̄ is a fixed
point of (3.42) if φ(x̄) = x̄. Then, the constant sequence, x(t) = x̄ for all
t ≥ 0, is a solution of (3.42). Fixed points are important, because, as we
argued in section 3.1, limiting values of solutions are fixed points.
Cobwebbing is carried out in the Cartesian plane. Start by graphing the
diagonal line y = x and, superimposed on that, the graph, y = φ(x), of the
functi0n φ appearing in (3.42). The fixed points of φ are then easy to read
off, since they are just the x-coordinates of the the points in the plane where
the graphs of y = φ(x) and y = x intersect. Figure 1 shows an example;
the unique fixed point in Figure 1 is labeled x̄. The object of cobwebbing
is to plot successive values of the solution to (3.42), starting at any given
initial condition x(0) = x0 , on the x-axis. The fundamental operation of
cobwebbing is a stepping procedure that, starting from any point (x, x) on
the diagonal, leads to the point (φ(x), φ(x)). Figure 1 shows how the method
works. Plot the initial point x(0) = x0 on the x-axis. First draw a vertical
line segment from the point (x0 , x0 ) to the curve y = φ(x), and then
draw a horizontal line from the curve back to y = x. The result looks
like a step. Since the vertical line intersects the graph of y = φ(x) at the
ordinate y = φ(x0 ), the horizontal line is drawn at the level y = φ(x0 ) and
will intersect the line y = x at (φ(x0 ), φ(x0 )). So the first graphical step
indeed leads from (x0 , x0 ) to (φ(x0 ), φ(x0 )) = (x(1), x(1)). Iteration of
the stepping procedure starting from (x(1), x(1)) then produces the point
(φ(x(1), φ(x(1)) = (x(2), x(2)); a third repetition leads to (x(3), x(3)), and
so on. Thus, the x-coordinates of the successive points on y = x hit by
the stepping procedure plot out the solution to equation (3.42). Figure 1
carries out the first few iterations.
Cobwebbing can help one see easily how solutions to a difference equation
behave starting from different initial values x0 . For example, it is clear from
Figure 1, that the successive values of the plotted solution x(t) will increase
and will converge to the fixed point x̄, as t → ∞. It is also clear that the
same limiting behavior will obtain for any initial values x0 close to but less
than x̄. If x0 is close to x̄ but larger than x̄, cobwebbing will show that the
36
CHAPTER 3. POPULATION GENETICS I
6
y=x
y
y = φ(x)
φ(x0 )
x0
x(1) x(2)x(3) x̄
-
x
Figure 1.
solution decreases toward x̄; you should check this. Although the argument
is not rigorous, it is clear that limt→∞ x(t) = x̄ for all starting values x0
sufficiently close to x̄. In the case of Figure 1, x̄ is an example of a stable fixed
point. In general, if x∗ is a fixed point, the set of x such that limt→∞ x(t) =
x∗ , where (x(t))t≥0 is the solution of x(t+1) = φ(x(t)) starting at x(0) = x,
is called the basin of attraction of x∗ . Then, x∗ is called stable if its basin
of attraction includes an open interval about x∗ .
Figure 2 illustrates a quite different situation for a different function η
around its fixed point x∗ . The cobwebbing is not shown, and you should
supply it yourself. You will see that when the initial point is close to x∗ ,
the successive points x(1), x(2), . . . of the solution to x(t+1) = η(x(t)) will
spiral away from x∗ . The picture this time really will look like a cobweb,
and x∗ is not stable.
There are nice sufficient conditions that a fixed point be either stable or
unstable. Assume that φ is differentiable at a fixed point x∗ . The tangent
line to the graph of φ at x∗ is defined by y = φ(x∗ ) + φ0 (x∗ )(x − x∗ ) =
x∗ + φ(x∗ )(x − x∗ ), and, for values of x close to x∗ , approximates the graph
of y = φ(x) nicely. Indeed, let e(x) = φ(x) − [x∗ + φ(x∗ )(x − x∗ )] denote the
3.4. AN INFINITE POPULATION MODEL WITH SELECTION
37
error made in approximating φ(x) by x∗ + φ(x∗ )(x − x∗ ). Then, as a direct
consequence of the definition of derivative,
lim e(x)/(x − x∗ ) = 0;
x→x∗
in words, the error is vanishingly small compared to |x − x∗ | as x approaches
x∗ . With this observation in mind, write the difference equation x(t+1) =
φ(x(t)) of (3.42) in the form
x(t+1) = x∗ + φ0 (x∗ )(x(t) − x∗ ) + e (x(t)) .
The equation obtained by dropping the error term is
x(t+1) = x∗ + φ0 (x∗ )(x(t) − x∗ ).
This is called the linearization of (3.42); it is a linear difference equation
with a fixed point at x∗ and it should be a good approximation to (3.42), as
long as x(t) is close to x∗ . In fact, when |φ0 (x∗ )| < 1, the results of section
3.1 imply that x∗ is a stable fixed point of the linearized equation, and,
using this, it can be shown x∗ is a stable point for (3.42) as well. Likewise,
when |φ(x∗ )| > 1, x∗ is unstable for the linear system, and hence also for
(3.42). The rigorous proof shall not be given here. We content ourselves
with stating the result formally in the following theorem.
6
y=x
y
y = η(x)
x∗
Figure 2.
x
-
38
CHAPTER 3. POPULATION GENETICS I
Theorem 2 Let z be a fixed point of a continuously differentiable function
φ. If
If | φ0 (z) |< 1, then z is a stable fixed point.
If | φ0 (z) |> 1,
then z is not stable.
We have given one plausibility argument for this theorem. The reader
should supply another by drawing a number of examples, some satisfying
|φ0 (z)| > 1, others satisfying |φ0 (z)| < 1, and graphing solutions by cobwebbing.
If |φ0 (x∗ )| = 1, the fixed point x∗ can be either stable or not stable;
analysis of the linearized equation cannot distinguish between the two cases.
3.4.2
Exercises
Exercise 3.4.1. Show that
√
2 is a stable fixed point of
x(t+1) =
x(t)
1
+
.
2
x(t)
(This equation appeared
in Example 3.1.2. Note in this example how close
√
already x(2) is to 2 when x(0) = 1. This difference equation is actually
Newton’s method for finding roots of x2 − 2.)
Exercis 3.4.2. Graph φ(x) = (x3√− 4x)/8 carefully. Explore the solutions
starting from x(0) = 1, x(0) = 2/ 3 and x(0) = 3. From cobwebbing, guess
the basin of attraction of the fixed point x∗ = 0. Show that your guess is
correct by a rigorous argument. (For this problem use a graphing calculator
or a mathematical package such as MAPLE or MATHEMATICA.)
Exercise 3.4.3. Consider the difference equation
x(t+1) = f (x(t)),
where f (x) = 4x(1 − x). Graph this function on the unit interval [0, 1] and
notice that f maps the unit interval into itself. A solution of period 2 to the
difference equation is a sequence of the form (z, w, z, w, . . .), where z 6= w;
that is, w = f (z) and z = f (w), so the solution alternates between these two
values. Argue by cobwebbing that it is plausible x(t+1) = 4x(t)(1 − x(t))
has a solution of period 2. Then find a solution of period 2 analytically;
determine exact values of z and w.
A period 2 solution is stable if for all x(0) close to z, the solution converges to the period 2 solution, in the sense that lims→∞ x(2s) = z and
lims→∞ x(2s + 1) = w. Show that the periodic solution you found is not
stable by considering what difference equation y(s) = x(2s) solves.
3.4. AN INFINITE POPULATION MODEL WITH SELECTION
3.4.3
39
A Model with Selection
In this section, we keep assumptions (A.1)—(A.4) of the basic model: we
study an infinite, monecious population with non-overlappling generations
and random mating. In addition, for concreteness, we assume that mating occurs seasonally; generation t produces generation t+1 all at one time
and then generation t + 1 matures over an interval of time until the next
mating season. Also, migration and mutation are not allowed. However, we
now assume that selection occurs because genotype affects the probability of
survival to reproductive maturity. The survival probabilities, which are also
called selection coefficients, will be denoted by wAA , wAa and waa , where, for
example, wAA denotes the probability that an individual of genotype AA
survives from birth to reproductive maturity. We suppose these survival
probabilities are the same from generation to generation and individuals
survive or don’t survive independently from one another. Survival is therefore equivalent probabilistically to the following experiment. Endow each
individual at birth with a coin appropriate to its genotype; an individual
with genotype AA gets a coin for which the probability of heads is wAA , an
Aa individual gets a coin for which the probability of heads is wAa , and so
on. Each individual flips its coin independently and survives to reproductive
maturity only if it flips heads.
The first goal is to derive a model from our assumptions for the usual
case of one locus with two alleles. For this it is necessary to define two
sets of genotype and allele frequencies per generation. The old notation,
fA (t), fAA (t), etc., will denote the frequencies in generation t at the time
of its birth. The notation pA (t), pAA (t), etc., will denote the frequencies in
generation t at the time of reproductive maturity when it is giving birth to
the next generation. Because the population is infinite and individuals survive independently of one another, the law of large numbers implies pAA (t)
equals the conditional probability that an individual has genotype AA given
that it survives, and similarly for pAa (t) and paa (t).
The derivation proceeds in two steps. The first step relates frequencies
between generations, the second within generations. The first step is easy
given what we know. The probability that a randomly chosen parent of
generation t passes allele A an offspring is pA (t) = pAA (t) + pAa (t)/2. Hence
by random mating and the infinite population assumption, for all t ≥ 0,
fA (t+1) = pA (t),
fAa (t+1) = 2pA (t)(1 − pA (t)),
fAA (t+1) = p2A (t),
(3.43)
2
faa (t+1) = (1 − pA (t))
The second step is to express the probabilities (pAA (t), pAa (t), paa (t)) in
terms of (fAA (t), fAa (t), faa (t)) and the selection coefficients. For example,
pAA (t) is the conditional probability that an individual has genotype AA
40
CHAPTER 3. POPULATION GENETICS I
given that it survives: in mathematical notation,
pAA (t) =
P (UAA ∩ S)
P (S|UAA )P (UAA )
=
P (S)
P (S)
where S is the event a randomly chosen individual born in generation t
survives and UAA is the event a randomly chosen individual is AA. By
definition, P (UAA ) = fAA (t) and P (S|UAA ) = wAA , and so the numerator
of pAA (t) is wAA fAA (t). As for the numerator, let UAa and Uaa denote the
event an individual born in generation t is Aa and, respectively aa. By the
law of total probabilities (see Chapter 2, Section 2.1.4),
P (S) = P (S|UAA )P (UAA ) + P (S|UAa )P (UAa ) + P (S|Uaa )P (Uaa )
= wAA fAA (t) + wAa fAa (t) + waa faa (t).
Thus,
pAA (t) =
wAA fAA (t)
.
wAA fAA (t) + wAa fAa (t) + waa faa (t)
If t ≥ 1, the equations of (3.43), applied with t − 1 replacing t, imply
generation t in infancy is in Hardy-Weinberg equilibrium with p = pA (t−1) =
fA (t); hence
pAA (t) =
wAA fA2 (t)
.
wAA fA2 (t) + wAa 2fA (t)(1 − fA (t)) + waa (1 − fA (t))2
(3.44)
The same reasoning shows that
pAa (t) =
wAA fA2 (t)
wAa 2fA (t)(1 − fA (t))
.
+ wAa 2fA (t)(1 − fA (t)) + waa (1 − fA (t))2
(3.45)
The final step of the derivation is simply to combine the results obtained
in equations (3.43), (3.44), and (3.45). From (3.43), fA (t + 1) = pA (t) =
pAA (t) + pAa (t)/2. Hence, adding (3.44) and one-half of (3.45), we find that
for t ≥ 1.
fA (t+1) =
wAA fA2 (t) + waA fA (t)(1 − fA (t))
.
wAA fA2 (t) + wAa 2fA (t)(1 − fA (t)) + waa (1 − fA (t))2
(3.46)
This equation will also be valid for t = 0, if we assume generation 0 in its
infancy is in Hardy-Weinberg equilibrium (that is, fAA (0) = fA2 (0), fAa (0) =
2fA (0)(1 − fA (0)), and faa (0) = (1 − fA (0))2 ), because (3.44) and (3.45) are
then valid for t = 0 also. From now on, let us impose this assumption. It
simplifies the model and does not affect the analysis of the limiting behavior
of fA (t).
3.4. AN INFINITE POPULATION MODEL WITH SELECTION
41
Equation (3.46), for all t ≥ 0, is thus the final model. It looks a little
scary, so, to simplify writing it, define the so-called fitness function
W (p) = p2 wAA + 2p(1 − p)wAa + (1 − p)2 waa ,
0 ≤ p ≤ 1.
Then the numerator in (3.46) is simply W (fA (t)), and we can write the
model as
fA (t+1) =
wAA fA2 (t) + waA fA (t)(1 − fA (t))
,
W (fA (t))
t ≥ 0.
(3.47)
Our derivation of this equation assumed that selection occurs because
different genotypes have different survival rates. But selection also can occur
because different genotypes have different probabilities of reproductive success, or because of a combination differential survival rates and reproductive
fitness. Fortunately, it is possible to reinterpret the selection coefficients, to
cover all these possibilities with the same model. The only assumption we
need is that the parents of a randomly chosen offspring of generation t are
chosen independently. We defined pAA (t) above was the frequency of AA in
generation t at the time of mating. However because random mating was
assumed, pAA (t) equals the probability that parent i has genotype AA, for
both parent i = 1 and parent i = 2, and this was really the only interpretation we ultimately used in the derivation of the final equation for fA (t).
Thus, for a more general model, in which we don’t worry exactly how selection occurs, let us interpret pAA (t), and similarly pAa (t) and paa (t), just
as genotype probabilities of parents in a mating. Then it will still be true,
as in equation (3.43), that fA (t+1) = pA (t) = pAA (t) + (pAa (t)/2). Next,
assuming wAa 6= 0, divide the expressions in equations 3.44) and (3.45); the
result is
wAA fAA (t)
pAA (t)
=
.
(3.48)
pAa (t)
wAa fAa (t)
A similar argument implies
waa faa (t)
paa (t)
=
.
pAa (t)
wAa fAa (t)
(3.49)
It turns out that these two equations, when employed in conjunction with
fA (t+1) = pA (t) = pAA (t)+(pAa (t)/2) and with pAA (t)+pAa (t)+paa (t) = 1,
also lead to equation (3.47) for (fA (t))t≥0 . Thus, we can recover the same
selection model by imposing (3.48) and (3.49) as assumptions that define the
selection coefficients. In this view, the selection coefficients are nonnegative
weights which quantify how the ratios of genotype probabilities in mating
differ from the genotype frequencies of the infant population. This approach
reveals that the model (3.47) depends only on the ratios of the selection
coefficients to each other. Therefore there are really only two free parameters
42
CHAPTER 3. POPULATION GENETICS I
among wAA , wAa , and waa in determining the model numerically. It is
common in the literature to parameterize the selection coefficient ratios using
two numbers r and s, by taking wAa = 1, wAA = 1−r and waa = 1−s. When
this parameterization is used, selection coefficients cannot be interpreted as
survival probabilities. Of course, this whole discussion assumed wAa > 0.
However, the equation (3.47) makes sense even when wAa = 0, and it can
be interpreted in the same general manner if one takes wAa = 0 to mean
that pAa = 0.
3.4.4
Analysis of the selection model
In this section, we use cobwebbing to analyse the selection model (3.47).
It shall be assumed that the selection coefficients wAA , wAa and waa are
all stricly positive, so that W (p) > 0 for all p in [0, 1]. For notational
convenience, f (t) will be used to denote fa (t), and φ(p) will denote the
function
p2 wAA + p(1 − p)wAa
.
φ(p) =
W (p)
Then, the difference equation (3.47) takes the form the form:
f (t+1) = φ(f (t)).
(3.50)
For any, strictly positive choice of the fitness coefficients, φ has fixed
points at 0 and 1; this is easy to check by direct calculation. These fixed
points make sense; p = 0 corresponds to the complete absence of allele A,
and if it is absent in one generation it cannot appear in future generations
because there is no mutation. Likewise, p = 1 corresponds to the complete
absence of allele a.
The graph of y = φ(p) will have one of four possible general shapes,
each corresponding to a different range of values of the selection coefficients.
These shapes are illustrated in Figures 4, 5, 6 and 7. The graphs are plotted
over the interval 0 ≤ p ≤ 1, which is the only region of interest—being a
frequency, the fA (t) must remain in the interval [0, 1] for all t. We shall
explain each graph, and its consequence for the behavior of solutions to
(3.50) on a case by case basis. The explanations require the following facts
about φ which are stated without proofs, which require only routine, if
somewhat messy, calculation.
First, φ has a third fixed point, found by looking for a solution p 6= 1 to
W (p) = pwAA + (1 − p)wAa , whenever 2wAa − wAA − waa 6= 0. It is
p̄ =
wAa − waa
.
2wAa − wAA − waa
(3.51)
This fixed point will satisfy 0 < p̄ < 1 if either wAa > wAA and wAa > waa ,
or wAa < wAA and wAa < waa , and in no other cases.
3.4. AN INFINITE POPULATION MODEL WITH SELECTION
43
Second, the derivative of φ is
φ0 (p) =
p2 wAA wAa + 2p(1 − p)wAA waa + (1 − p)2 wAa waa
.
W 2 (p)
(3.52)
This is always positive in the interval 0 ≤ p ≤ 1, and hence φ is always
strictly increasing in this interval.
y=x
6
y=x
6
-
Figure 4.
-
1
Figure 5.
y=x
6
y=x
6
-
-
p̄
Figure 6.
1
1
p̄
Figure 7.
1
Case I. Allele a is favored by selection: wAA < wAa < waa .
In this case, the graph of φ will have the shape shown in Figure 4. The
analsyis will show that if 0 ≤ f (0) < 1, then
lim f (t) = 0.
t→∞
(3.53)
This makes sense. It says that allele A will disappear from the population
if allele a has a selective advantage.
To see why Figure 4 is the correct graph, first use the fact, stated above,
that when wAA < wAa < waa , the fixed point p̄ does not lie in [0, 1]. Hence
the only fixed points of φ in 0 ≤ p ≤ 1 are p = 0 and p = 1, and the graph of
φ must lie either entirely above or entirely below the diagonal on the interval
44
CHAPTER 3. POPULATION GENETICS I
0 < p < 1. However, the slope of the tangent line to y = φ(p) at p = 0 is, by
equation (3.52), φ0 (0) = wAa /waa . Since 0 < wAa < waa , φ0 (0) < 1, which
implies the graph of φ must lie below the diagonal. Thus φ(p) < p for all
0 < p < 1.
Now pick a point f (0) between 0 and 1, but strictly less than 1 and
start cobwebbing. In every iteration, f (t+1) = φ(f (t)) < f (t). Therefore
successive values of f (t) decrease and can only tend to 0. This proves (3.53).
Case II. Allele A is favored by selection:
waa < wAa < wAA .
In this case, if 0 < f (0) ≤ 1,
lim f (t) = 1.
t→∞
This is really Case I with the roles of A and a reversed. The graph of φ
is shown in Figure 5. This time it lies stricly above the line y = p, and you
can convince yourself by cobwebbing that all solutions, except the solution
which starts and stays at p = 0, converge to 1.
Case III. Heterozygote dominance.
waA > wAA and waA > waa .
In this case, if 0 < f (0) < 1,
lim f (t) = p̄,
t→∞
where p̄ is the frequency defined above in (3.51).
The graph of φ for this case is shown in Figure 6. From the remark
after equation (3.51), we know the fixed point p̄ is strictly inside the interval
[0, 1]. Since φ0 (0) = waA /waa > 1, the graph of φ will be above the graph
of y = p for 0 < p < p̄. It will pass through y = p at p = p̄ and be below
y = p for p̄ < p < 1. The graph of y = φ(p) is always increasing. Thus, at
p̄, 0 < φ0 (p̄) < 1, and p̄ is a stable fixed point. By cobwebbing you can see
that whether f (0) is above or below p̄, limt→∞ f (t) = p̄.
This result is also very intuitive. It says that if heterozygotes are favored
by selection, both A and a alleles will be maintained in the population. The
fixed point p̄ provided by the model quantifies the ultimate balance between
the alleles.
Case IV. Homozygote dominance
waA < wAA and waA < waa .
In this case,
if 0 < f (0) < p̄, then lim f (t) = 0, and if p̄ < f (0) < 1, lim f (t) = 1.
t→∞
t→∞
(3.54)
The graph of y = φ(p) in this case is shown in Figure 7. The fixed point
p̄ is again inside [0, 1], but this time, the graph of y = φ(p) is below y = p
3.4. AN INFINITE POPULATION MODEL WITH SELECTION
45
for 0 < p < p̄ and above for p̄ < p < 1. Thus, the slope φ0 (p̄) of φ at p̄ is
strictly greater that 1, which implies that p̄ is not stable. The student can
check the validity of the limits stated in (3.54) by cobwebbing.
The interpretation of this case is also clear. If both homozygotes are
favored by selection over heterozygote, the population will eliminate heterozygosity by eliminating either allele A (f (t) → 0) or allele a (f (t) → 1).
But which allele wins depends on the initial frequency of A’s versus a’s and
the exact values of the selection coefficients. The value p̄ is a quantitative
expression for the boundary between the region in which a wins and that in
which A wins.
The analysis of this section shows that, despite the complex nonlinearity
of the selection model, it is not too difficult to analyze. The conclusions of
the analysis are all what one would expect intuitively. This could be grounds
for criticism. What use is all the work of modeling if the end result only
confirms what we know intuitively must be true? However, the model also
give quantitative information. In the case of heterozygote dominance it tells
us exactly what the limiting allele frequency must be, and in the case of
homozygote dominance, where the dividing line between the regions where
A takes over and a takes over lies.
3.4.5
Mean Fitness Increases
This section presents another perspective on how solutions to equation (3.50)
for the allele frequency f (t) evolve. It is sometimes called the Fundamental
Theorem of Natural Selection. Recall that we have called W the fitness
function. In the interpretation of the selection coefficients as survival probabilities, it was shown that W (fA (t)) is the probability that a randomly
selected individual from the infant population of generation t survives. This
justifies interpreting W (fA (t)) as the mean fitness of generation t.
Theorem 3 For the one locus/two allele selection model (3.47), mean fitness always increases from generation to generation. That is, for any t
W (f (t+1)) > W (f (t))
if f (t) is not a fixed point.
Comment: The mean fitness W provides what is called a Lyapunov function for the difference equation (3.50). Lyapunov functions for a dynamical
system are functions which are either increasing or decreasing along solutions, and they are very useful in analyzing the solution behavior. In the
case of the selection model, as t → ∞, the solution x(t) approaches a value
p in [0, 1] at which the fitness function achieves a local maximum, unless the
solution is at a fixed point.
46
CHAPTER 3. POPULATION GENETICS I
To prove Theorem 3) it is only necessary to show W (φ(p)) > W (p),
whenever p is a point in (0, 1) and p is not a fixed point. If this is true, then
W (f (t + 1)) = W (φ(f (t)) > W (f (t)), so long as f (t) is not a fixed point,
and this shows mean fitness increases.
The proof is just a computation. Plug φ(p) into W and calculate. The
result, after some messy computation, is
W (φ(p)) − W (p) =
(φ(p) − p)2
[W (p) + pwAA + (1 − p)waa ] .
p(1 − p)
(3.55)
This is strictly positive for every p in the interval (0, 1) such that φ(p) 6= p.
3.4.6
Problems
Exercise 3.4.4. Consider the study of a locus with two alleles A and a.
Assume that the selection coefficients have been determined to be wAA =
0.5, waa = 0.6 and wAa = 0.4 If at time t = 0, the genotype frequencies
are fAA (0) = 0.4, fAa (0) = 0.2 and faa (0) = 0.4, determined the limiting
frequencies of allele A in generation t as t tends to ∞.
Exercise 3.4.5. Imagine an isolated population that has had a chance to
evolve over a long period of time. Suppose that observations over many
generations show that it in every generation the probability of being born
AA is 0.16, of being born Aa is 0.48, and of being born aa is 0.36. It is
known that selection acts and that the survival probabilities of AA and aa
are wAA = 0.1 and waa = 0.2. What is the selection coefficient wAa for the
heterozygote genotype? (Hint: Assume that the stable genotype frequencies
represent the limit of a model with selection.)
Exercise 3.4.6. Suppose it is known that wAA = wAa = w and that w > waa .
This case was not actually covered in the text. Determine limt→∞ pA (t) by
analyzing the shape of φ and invoking cobwebbing.
(Hint: By calculating p̄ show that the only fixed points of φ in [0, 1] are
0 and 1. Calculate the value of φ0 (0)—see equation (3.52)—and use this and
knowledge about the fixed points to graph the general shape of φ.)
Exercise 3.4.7. Derive the formula for the fixed point p̄ in (3.51). Derive
the formula given in the text for φ0 (p).
Exercise 3.4.8. Derive formula (3.55) in the proof that mean fitness increases.
Exercise 3.4.9. Suppose that in addition to selection, allele A mutates to a
with probability u and a to A with probability v in the course of mating.
Derive a difference equation for fA (t).
3.5. NOTES AND REFERENCES
3.5
47
Notes and References
1. Finite difference equation. The first order difference of a sequence
{x(t)} is the sequence {x(t + 1) − x(t)}. The term first order difference
equation is most properly applied to equations of the form
x(t+1) − x(t) = ψ(x(t)).
But, as is standard, we have used the term to refer to any equation of the
sort x(t+1) = φ(x(t)); of course this equation can be written in the form
x(t+1)−x(t) = φ(x(t))−x(t) so that it truly contains a first oreder difference
{x(t)}, but this is rather artificial. Difference equations, as we have defined
them, are really examples of discrete-time dynamical systems.
Finite difference equations occur throughout mathematics, often as the
expression of an algorithm. For example, Newton’s method for finding a root
of the function f is x(t+1) − x(t) = f (x(t))/f 0 (x(t)). Euler’s method for
approximating the solution of the first order differential equation x0 = g(x)
is x(t+h) − x(t) = hg(x(t)). The popular autoregressive moving average
processes for the analysis of time series are finite difference equation models.
A source for traditional theory of difference equations is Kenneth S. Miller,
An Introduction to the Calculus of Finite Differences and Difference Equations, Dover Publications, New York, 1966.
Mathematical ecologists and epidemiologists model many biological phenomena—predator-prey models, population growth, etc.— with difference
equations. In fact, it was the mathematical ecologist Robert May who first
studied how solutions to the discrete logistic equation,
x(t+1) = λx(t) (1 − x(t)) ,
depend upon the parameter λ. He published his first work on this equation
in the journal Nature, volume 261, pages 459-467, 1976. In this study he
discovered the phenomenon of chaos, that is, sensitive dependence on initial
conditions, for certain ranges of values of λ. May’s work was an important
inspiration to the development of the popularly known theory of chaos in
dynamical systems. The behavior of solutions to even very simple families of difference equations can be very rich. The textbook, K.T. Alligood,
T.D. Sauer, J.A. Yorke, Chaos, Springer-Verlag, New York, 1996, is one
among several introductory-level books on the subject.
2. The population genetics models presented here are all standard. I have
been guided in my treatment by the following sources
• W.J. Ewens, Population Genetics, Methuen & Co., Ltd., London,
1969.
48
CHAPTER 3. POPULATION GENETICS I
• J.C. Kingman, Mathematics of Genetic Diversity, CBMS-NSF regional
conference series in applied math 34, SIAM, Philadelphia, 1980.
• S. Tavaré, Ancestral Inference in Population Genetics, in Lectures on
Probability Theory and Statistics; editor, J. Picard, Lecture Notes in
Mathematics 1837, Springer-Verag, Berlin, 2004.
• D.L. Hartl and A.G. Clark, Principles of Population Genetics, second
edition, Sinauer Associates, Sunderland, MA, 1989.
The first three sources are at a mathematical level higher than this text.
The third book is a standard population genetics text covering the scientific
issues and presenting data, as well as the math.